Question 15 Marks
$2 kg$ of ice melts when water at $100^{\circ} C$ is poured in a hole drilled in a block of ice. What mass of
water was used? Given: Specific heat capacity of water $=$ $4200 Jkg ^{-1} K ^{-1}$, Lice $=336 \times 10^3 Jkg ^{-1}$.
Answer>Since the whole block does not melt and only 2 kg of it melts, so the final temperature would be >0 °C. >Amount of heat energy gained by 2 kg of ice at 0°C to convert into water at 0°C = 2 × 336000 = 672000 J >Let amount of water poured = m kg. >Initial temperature of water = 100°C. >Final temperature of water = 0°C. >Amount of heat energy lost by m kg of water at 100oC to reach temperature 0°C = m × 4200 × >100 = 420000m J >We know that heat energy gained =heat energy lost. >672000J = m × 420000J >m = 672000/420000 = 1.6kg
View full question & answer→Question 25 Marks
$250 g$ of water at $30^{\circ} C$ is contained in a copper vessel of mass $50 g$.
Calculate the mass of ice
required to bring down the temperature of the vessel and its contents to $5^{\circ} C$. Given: specific
latent heat of fusion of ice $=336 \times 10^3 Jkg ^{-1}$, specific heat capacity of copper $=400 Jkg ^{-1} K ^{-1}$,
specific heat capacity of water $=4200 Jkg ^{-1} K ^{-1}$
AnswerMass of copper vessel $m _1=50 g$.
Mass of water contained in copper vessel $m _2=250 g$.
Mass of ice required to bring down the temperature of vessel $= m$
Final temperature $=5^{\circ} C$. Amount of heat gained when ' m ' g of ice at $0^{\circ} C$ converts into water at $0^{\circ} C = m \times 336 J$
Amount of heat gained when temperature of ' m ' g of water at $0^{\circ} C$ rises to $5^{\circ} C = m \times 4.2 \times 5$
Total amount of heat gained $= m \times 336+ m \times 4.2 \times 5$
Amount of heat lost when 250 g of water at $30^{\circ} C$ cools to $5^{\circ} C =250 \times 4.2 \times 25=26250 J$
Amount of heat lost when 50 g of vessel at $30^{\circ} C$ cools to $5^{\circ} C =50 \times 0.4 \times 25=500 J$
Total amount of heat lost $= 26250 + 500 = 26750 J$
We know that amount of heat gained = amount of heat lost
$m \times 336 + m \times 4.2 \times 5 = 26750$
$357 m = 26750$
$m = 26750/357 = 74.93 g$
Hence, mass of ice required is $74.93 g.$
View full question & answer→Question 35 Marks
A piece of ice of mass $40 g$ is added to $200 g$ of water at $50^{\circ} C$ Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = $4200 Jkg ^{-1} K ^{-1}$, and specific latent heat of fusion of ice $=$ $336 \times 10^3 Jkg ^{-1}$.
AnswerLet final temperature of water when all the ice has melted $= T ^{\circ} C$. Amount of heat lost when $200 g$ of water at $50^{\circ} C$ cools to $T ^{\circ} C =$ $200 \times 4.2 \times(50-T)=42000-840 T$
Amount of heat gained when $40 g$ of ice at $0 oC$ converts into water at OoC. $=40 \times 336 J =13440 J$
Amount of heat gained when temperature of $40 g$ of water at $0 oC$ rises to $ToC =40 \times 4.2 \times( T -0)=168 T$
We know that
Amount of heat gained = amount of heat energy lost.
$
\begin{aligned}
& 13440+168 T =42000-840 T \\
& 168 T +840 T =42000-13440 \\
& 1008 T =28560 \\
& T =28560 / 1008=28.33^{\circ} C .
\end{aligned}
$
View full question & answer→Question 45 Marks
Find the result of mixing $10 g$ of ice at $-10^{\circ} C$ with $10 g$ of water at $10^{\circ} C$. Specific heat capacity
of ice $=2.1 Jkg ^{-1} K ^{-1}$, Specific latent heat of ice $=336 Jg ^{-1}$ and specific heat capacity of water
$
=4.2 Jkg ^{-1} K ^{-1}
$
AnswerLet whole of the ice melts and let the final temperature of the mixture be T°C.
Amount of heat energy gained by 10g of ice at -10°C to raise its temperature to 0°C = 10 × 10 ×
2.1 = 210J
Amount of heat energy gained by 10g of ice at 0°C to convert into water at 0°C = 10 × 336 =
3360 J
Amount of heat energy gained by 10g of water (obtained from ice) at 0oC to raise its temperature
to T°C = 10 × 4.2 × (T − 0) = 42T
Amount of heat energy released by 10g of water at 10°C to lower its temperature to T°C = 10
4.2 × (10 − T) = 420 − 42T
Heat energy gained = Heat energy lost
210 + 3360 + 42T = 420 − 42T
T = −37.5°C
This cannot be true because water cannot exist at this temperature.
So whole of the ice does not melt. Let m gm of ice melts. The final temperature of the mixture
becomes 0°C.
So, amount of heat energy gained by 10g of ice at -10°C to raise its temperature to 0°C = 10 ×10 × 2.1 = 210J
Amount of heat energy gained by m gm of ice at 0°C to convert into water at 0°C =m × 336 =
336m J
Amount of heat energy released by 10g of water at 10°C to lower its temperature to 0°C = 10 ×4.2 × 10 − 0) = 420
Heat energy gained = Heat energy lost
210 + 336m = 420
m = 0.625 gm
View full question & answer→Question 55 Marks
In an experiment, $17 g$ of ice is used to bring down the temperature of $40 g$ of water at $34^{\circ} C$ to its freezing temperature. The specific heat capacity of water is $4.2 Jkg ^{-1} K ^{-1}$. Calculate the specific latent heat of ice. State one important assumption made in the above calculation.
AnswerMass of ice $m 1=17 g$
Mass of water $m 2=40 g$.
Change in temperature $=34-0=34 K$
Specific heat capacity of water is $4.2 Jg ^{-1} K ^{-1}$.
Assuming there is no loss of heat, heat energy gained by ice (latent
heat of ice), Q= heat energy
released by water
$
Q =40 \times 34 \times 4.2=5712 J
$
Specific latent heat of ice $=$
$
L=\frac{Q}{m}=\frac{5712}{17}=336 J^{-1}
$
View full question & answer→Question 65 Marks
A refrigerator converts $100 g$ of water at $20^{\circ} C$ to ice at $-10^{\circ} C$ in 73.5 min. Calculate the average
the rate of heat extraction in watt. The specific heat capacity of water is $4.2 Jkg ^{-1} K ^{-1}$ Specific latent heat of ice is $336 J g -1$ and the specific heat capacity of ice if $2.1 Jkg ^{-1} K ^{-1}$.
AnswerAmount of heat released when $100 g$ of water cools from $20 o$ to $0^{\circ} C$
$
=100 \times 20 \times 4.2=8400 J \text {. }
$
Amount of heat released when $100 g$ of water converts into ice at $0^{\circ} C$
$
=100 \times 336=33600 J .
$
Amount of heat released when $100 g$ of ice cools from $0 oC$ to $-10 oC$
$
=100 \times 10 \times 2.1=2100 J \text {. }
$
Total amount of heat $=8400+33600+2100=44100 J$.
Time taken $=73.5 min =4410 s$.
Average rate of heat extraction (power)
$
p=\frac{E}{t}=\frac{44100}{4410}=10 W
$
View full question & answer→Question 75 Marks
How much heat energy is released when 5.0 of water at $20^{\circ} C$ changes into ice at $0^{\circ} C$ ? Take
specific heat capacity of water $=4.2 Jkg ^{-1} K ^{-1}$, Specific latent heat of fusion of ice $336 Jg ^{-1}$
AnswerMass of water $m =5.0 g$
specific heat capacity of water $c =4.2 Jg ^{-1} K ^{-1}$
specific latent heat of fusion of iceL $=336 Jg ^{-1}$
Amount of heat energy released when $5.0 g$ of water at $20^{\circ} C$ changes into water at $0^{\circ} C =5 \times 4.2$
$
\times 20=420 J \text {. }
$
Amount of heat energy released when $5.0 g$ of water at $0^{\circ} C$ changes into ice at $0^{\circ} C =5 \times 336 J =$
$
1680 J \text {. }
$
Total amount of heat released $=1680 J +420 J =2100 J$.
View full question & answer→Question 85 Marks
$10 g$ of ice at $0^{\circ} C$ absorbs $5460 J$ of heat energy to melt and change to water at $50^{\circ} C$. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is $4200 J kg ^{-1} K^{-1}$.
AnswerGiven, Mass of ice $=10 g$
Heat energy absorbed $(Q)=5460 J$
Specific heat capacity of water $=4200 J kg ^{-1} K ^{-1}=4.2 J g ^{-1} K ^{-1}$
Specific latent heat of fusion of ice $=$ ?
(i) Heat energy required to melt the ice at $0^{\circ} C$ to water at $0^{\circ} C \left( Q _1\right)=$ $m \times L$
Substituting the values in the formula we get,
$Q _1=10 \times L$
(ii) Heat energy required to raise temperature from $0^{\circ} C$ to $50^{\circ} C = m$ $\times c \times$ rise in temperature
Substituting the values in the formula we get,
$ Q_2=10 \times 4.2 \times 50$
$ Q_2=2100 J$
Then,
$ Q=Q_1+Q_2$
$\therefore 5460 J =10 \times L +2100 J$
$\therefore 5460 J -2100 J =10 \times L$
$\therefore 3360=10 \times L$
$\therefore \frac{3360}{10}= L$
$\therefore L =336 Jg ^{-1}$
View full question & answer→Question 95 Marks
$200 g$ of ice at $0^{\circ} C$ converts into water at $0^{\circ} C$ in 1 minute when heat is supplied to it at a constant rate. In how much time, $200 g$ of water at $0^{\circ} C$ will change to $20^{\circ} C$ ? Take specific latent heat of ice $=336 Jg ^{-1}$.
AnswerMass of ice, mice $=200 g$
Time for ice to melt, $t 1=1 min =60 s$
Mass of water, $mw =200 g$
Temperature change of water, $\Delta T =20^{\circ} C$
Rate of heat exchange is constant. So, power required for converting ice to water is same as the power required to increase the temperature of water.
$\therefore P_{\text {ice }}=P_{\text {water }}$
$\therefore \frac{E_{\text {ice }}}{t_1}=\frac{E_{\text {water }}}{t_2}$
$\therefore \frac{ m _{\text {ice }} L }{ t _1}=\frac{ m _{ w } C _{ w } \Delta T }{ t _2}$
$\therefore t _2=\frac{ m _{ w } C _{ w } \Delta T \times t _1}{ m _{\text {ice }} L }$
$\therefore t _2=\frac{200 \times 4.2 \times 20 \times 60}{200 \times 336}$
$\therefore t _2=15 s $
Hence, the time required is $15$ seconds.
View full question & answer→Question 105 Marks
The amount of heat energy required to convert $1 kg$ of ice at $-10^{\circ} C$ to water at $100^{\circ} C$ is $7,77,000 J$. Calculate the specific latent heat of ice. Specific heat capacity if ice $=2100 Jkg ^{-1} K ^{-1}$, Specific heat capacity of water $=4200 Jkg ^{-1} K ^{-1}$
AnswerAmount of heat energy gained by $1 kg$ of ice at $-10^{\circ} C$ to raise its temperature to $0^{\circ} C =1 \times 2100 \times 10=21000 J$
Amount of heat energy gained by $1 kg$ of ice at $0^{\circ} C$ to convert into water at $0^{\circ} C = L$
Amount of heat energy gained when temperature of $1 kg$ of water at $0 \circ C$ rises to $100^{\circ} C =1 \times 4200 \times 100=420000 J$
Total amount of heat energy gained $=21000+420000+L=441000$ $+ L$.
Given that total amount of heat gained is $=777000 J$.
So,
$
\begin{aligned}
& 441000+ L =777000 \\
& L =777000-441000 \\
& L=336000 J K g^{-1}
\end{aligned}
$
View full question & answer→Question 115 Marks
Calculate the total amount of heat energy required to convert $100 g$ of ice at $-10^{\circ} C$ completely into water at $100^{\circ} C$. Specific heat capacity of ice $=2.1 kg ^{-1} K ^{-1}$, specific heat capacity of water $=$ 4.2 $Jkg ^{-1} K ^{-1}$, specific latent heat of ice $=336 Jg ^{-1}$
AnswerAmount of heat energy gained by $100 g$ of ice at $-10^{\circ} C$ to raise its temperature to $0^{\circ} C =$
$
100 \times 2.1 \times 10=2100 J
$
Amount of heat energy gained by $100 g$ of ice at $0^{\circ} C$ to convert into water at $0^{\circ} C =$
$
100 \times 336=33600 J
$
Amount of heat energy gained when temperature of $100 g$ of water at $0^{\circ} C$ rises to $100^{\circ} C =$
$
100 \times 4.2 \times 100=42000 J
$
Total amount of heat energy gained is $=2100+33600+42000=$ $77700 J =7.77 \times 10^4 J$
View full question & answer→Question 125 Marks
How does the average potential energy of molecules of a substance change during its change in phase at a constant temperature?
AnswerAverage potential energy increases.
Explanation : When a substance is heated at constant temperature (i.e. during its phase change state), the heat supplied makes the vibrating molecules gain potential energy to overcome the intermolecular force of attraction and move about freely. This means that the substance changes its form.
However, this heat does not increase the kinetic energy of the molecules, and hence, no rise in temperature takes place during the change in phase of a substance.
This heat supplied to the substance is known as latent heat and is utilized in changing the state of matter without any rise in temperature.
View full question & answer→Question 135 Marks
Water in lakes and ponds do not freeze at once in cold countries. Give reason.
AnswerThe specific latent heat of fusion of ice is sufficiently high, about $336 J g^{-1}.$ Before freezing the water in the lakes and ponds will have to release a large quantity of heat to the surrounding. If there is any layer of ice formed on water than water being a poor conductor of heat will also prevent the loss of heat from water of lake. Hence, in cold countries water in lakes and ponds do not freeze.
View full question & answer→Question 145 Marks
An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0 ℃ to
15.0 ℃ I 100 s. Calculate: (i) the heat capacity of 4.0 kg of liquid, (ii) the specific heat capacity
of the liquid.
AnswerPower of heater $P =600 W$
Mass of liquid $m=4.0 kg$
Change in temperature of liquid $=(15-10)^{\circ} C =5^{\circ} C$ (or $5 K$ )
Time taken to raise its temperature $=100 s$
Heat energy required to heat the liquid
$\Delta Q=m c \Delta T$
And
$
\Delta Q=P \times t=600 \times 100=60000 J
$
$c=\frac{\Delta Q}{m \Delta T}=\frac{60000}{4 \times 5} 3000 J K g^{-1} K^{-1}$
Heat capacity $= c \times m$
Heat capacity $=4 \times 3000 JKg ^{-1} K ^{-1}=1.2 \times 10^4 J / K$
View full question & answer→Question 155 Marks
Find the time taken by a $500 W$ heater to raise the temperature of 50 $kg$ of material of specific
heat capacity $960 J kg ^{-1} K ^{-1}$, from $18^{\circ} C$ to $38^{\circ} C$. assume that all the heat energy supplied by the heater is given to the material.
AnswerSpecific heat capacity of material c $=960 J kg ^{-1} K^{-1}$
Change in temperature $\Delta T =(38-18)^{\circ} C =20^{\circ} C$ (or $20 K$ )
Power of heater $P =500 W$
$
\begin{aligned}
& \Delta Q=m c \Delta T \\
& \Delta Q=50 \times 960 \times 20
\end{aligned}
$
Time taken by a heater to raise the temperature of material
$\begin{aligned} & t =\frac{\Delta Q }{ P } \\ & t =\frac{50 \times 960 \times 20}{500} \\ & t =1920 \text { seconds } \\ & t =32 min \end{aligned}$
View full question & answer→Question 165 Marks
1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20° C to 40°C.
Calculate the specific heat capacity of lead
AnswerHeat energy supplied $=1300 J$
Mass of lead $=0.5 kg$
Change in temperature $=(40-20)^{\circ} C =20^{\circ} C$ (or $\left.20 K \right)$
Specific heat capacity of lead
$\begin{aligned} & c=\frac{\Delta Q}{m \Delta T} \\ & c=\frac{1300}{0.5 \times 20} \\ & c=130 J K g^1 K^1\end{aligned}$
View full question & answer→Question 175 Marks
(a) Calculate the heat capacity of a copper vessel of mass $150 g$ if the specific heat capacity of copper is $410 Jkg ^{-1} K ^{-1}$
(b) How much heat energy will be required to increase the temperature of the vessel in part (a) from $25^{\circ} C$ to $35^{\circ} C$ ?
Answer$
\begin{aligned}
& \text { Mass of copper vessel }=150 g \\
& =0.15 kg
\end{aligned}
$
The specific heat capacity of copper $=410 Jkg ^{-1} K ^{-1}$.
Heat capacity $=$ Mass $\times$ specific heat capacity
$
\begin{aligned}
& =0.15 kg \times 410 Jkg ^{-1} K ^{-1} \\
& =61.5 JK
\end{aligned}
$
Change in temperature $=(35-25)^{\circ} C =10^{\circ} C =10 K$
(ii) Energy required to increase the temperature of vessel
$
\begin{aligned}
& \Delta Q=m c \Delta T \\
& =0.15 \times 410 \times 10 \\
& =615 J
\end{aligned}
$
View full question & answer→Question 185 Marks
$1.0 kg$ of water is container in a $1.25 kW$ kettle. Calculate the time taken for the temperature of water to rise from $25^{\circ} C$ to its boiling point $100^{\circ} C$. Specific heat capacity of water $=4.2 J g ^{-1} K ^{-1}$.
Answer$
m =1.0 kg ; \Delta t =100-25=75^{\circ} C
$
Specific heat capacity $=4.2 J g ^{-1} k ^{-1}=4200 J kg ^{-1} k ^{-1}$
Power of kettle $=1.25 kw =1250$ watt
Let time be t second,
$\varepsilon= P \times t$ and $\theta= mc \Delta t$
$\therefore P \times t = mc \Delta t$
$\therefore 1250 \times t=1 \times 4200 \times 75$
$\therefore t =\frac{4200 \times 75}{1250}$
$
\therefore t =84 \times t
$
$
\therefore t =252 sec
$
$
\therefore t =\frac{252}{60 \min .}
$
$\therefore t =4.2 min$.
$
\therefore t =4 min 2 sec
$
View full question & answer→Question 195 Marks
The temperature of 600 g of cold water rises by 15° C when 300 g of hot water at 50° C is added to it. What was the initial temperature of the cold water?
AnswerMass of hot water $\left(m_1\right)=300 g$
Temperature $\left(T_1\right)=50^{\circ} C$
Mass of cold water $\left(m_2\right)=600 g$
Change in temperature of cold water $\left(T-T_2\right)=15^{\circ} C$
Final temperature $= T ^{\circ} C$
The specific heat capacity of water is $C$.
$m_1 c_1\left(T_1-T\right)=m_2 c_2\left(T-T_2\right)$
$300(50-T)=600(15)$
$T =20^{\circ} C$
Final temperature $=20^{\circ} C$
Change in temperature $=15^{\circ} C$
Initial temperature of cold water $=20^{\circ} C -15^{\circ} C =5^{\circ} C$.
View full question & answer→Question 205 Marks
$200 g$ of hot water at $80^{\circ} C$ is added to $300 g$ of cold water at $10^{\circ} C$. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water $=4200 Jkg ^{-1} K ^{-1}$
AnswerMass of hot water $\left(m_1\right)=200 g$
Temperature of hot water $\left(T_1\right)=80^{\circ} C$
Mass of cold water $\left( m _2\right)=300 g$
Temperature of cold water $\left(T_2\right)=10^{\circ} C$
Final temperature $(T)=$ ?
$\begin{aligned} & m_1 c_1\left(T_1-T\right)=m_2 c_2\left(T-T_2\right) \\ & c_1=c_2 \\ & T=\frac{m_1 T_1+m_2 T_2}{m_2+m_1} \\ & T=\frac{200 \times 80+300 \times 10}{500} \\ & T =38^{\circ} C \end{aligned}$
View full question & answer→Question 215 Marks
$45 g$ of water at $50^{\circ} C$ in a beaker is cooled when $50 g$ of copper at $18^{\circ} C$ is added to it. The contents are stirred till a final constant temperature is reached. Calculate this final temperature.The specific heat capacity of copper is $0.39 Jg ^{-1} K ^{-1}$ and that of water is 4.2 $Jg ^{-1} K ^{-1}$ State theassumption used.
AnswerMass of water $\left(m_1\right)=45 g$
Temperature of water $\left(T_1\right)=50^{\circ} C$
Mass of copper $\left(m_2\right)=50 g$
temperature of $\operatorname{copper}\left( T _2\right)=18^{\circ} C$
Final temperature $(T)=$ ?
The specific heat capacity of the copper $c_2=0.39 J / g / K$
The specific heat capacity of water $c_1=4.2 J / g / K$
$m_1 c_1\left(T_1-T\right)=m_2 c_2\left(T-T_2\right)$
$T=\frac{m_1 c_1 T_1+m_2 c_2 T_2}{m_2 c_2+m_1 c_1}$
$T=\frac{45 \times 4.2 \times 50+50 \times 0.39 \times 18}{45 \times 4.2+50 \times 0.39}=\frac{9801}{208.5}=47 c$
$T =47 C$
View full question & answer→Question 225 Marks
Discuss the role of high specific heat capacity of water with reference to climate in coastal areas.
AnswerThe specific heat capacity of water is very high. It is about five times as high as that of sand.
Hence the heat energy required for the same rise in temperature by a certain mass of water will
be nearly five times than that required by the same mass of sand. Similarly, a certain mass of
water will give out nearly five times more heat energy than that given by sand of the same mass
for the same fall in temperature. As such, sand gets heated or cooled more rapidly as compared
to water under the similar conditions. Thus a large difference in temperature is developed
between the land and the sea due to which land and sea breezes are formed. These breezes make
the climate near the sea shore moderate.
View full question & answer→Question 235 Marks
Why do the farmers fill their fields with water on a cold winter night?
AnswerIn the absence of water, if on a cold winter night the atmospheric temperature falls below 0oC,
the water in the fine capillaries of plant will freeze, so the veins will burst due to the increase in
the volume of water on freezing. As a result, plants will die and the crop will be destroyed. In
order to save the crop on such cold nights, farmers fill their fields with water because water has
high specific heat capacity, so it does not allow the temperature in the surrounding area of plants
to fall up to 0oC.
View full question & answer→Question 245 Marks
A mass m1 of a substance of specific heat capacity c1 at temperature t1 is mixed with a mass m2
of other substance of specific heat capacity c2 at a lower temperature t2. Deduce the expression
for the temperature of the mixture. State the assumption made, if any.
AnswerA mass $m_1$ of a substance $A$ of specific heat capacity $c_1$ at temperature $T_1$ is mixed with a mass
$m_2$ of other substance $B$ of specific heat capacity $c_2$ at a lower temperature $T_2$ and final
temperature of the mixture becomes $T$.
Fall in temperature of substance $A=T_1-T$
Rise in temperature of substance $B = T - T _2$
Heat energy lost by $A=m_1 \times c_1 \times$ fall in temperature
$
= m _1 c _1\left( T _1- T \right)
$
Heat energy gained by $B=m_2 \times c_2 \times$ rise in temperature
$
= m _2 C _2\left( T - T _2\right)
$
If no energy lost in the surrounding, then by the principle of mixtures, Heat energy lost by $A =$ Heat energy gained by $B$
$
m_1 c_1\left(T_1-T\right)=m_2 c_2\left(T-T_2\right)
$
After rearranging this equation, we get
$T=\frac{m_1 c_1 T_1+m_2 c_2 T_2}{m 1 c 1+m_2 c_2}$
Here we have assumed that there is no loss of heat energy.
View full question & answer→Question 255 Marks
Name three factors on which heat energy absorbed by a body depends and state how does it
depend on them.
AnswerThe quantity of heat energy absorbed by a body depends on three factors :
(i)Mass of the body - The amount of heat energy required is directly proportional to the mass of
the substance.
(ii)Nature of material of the body - The amount of heat energy required depends on the nature
on the substance and it is expressed in terms of its specific heat capacity c.
(iii)Rise in temperature of the body - The amount of heat energy required is directly proportional
to the rise in temperature.
View full question & answer→Question 265 Marks
Differentiate between heat capacity and specific heat capacity.
AnswerHeat capacity of the body is the amount of heat required to raise the temperature of (whole) body
by $1^{\circ} C$ whereas specific heat capacity is the amount of heat required
to raise the temperature of unit mass of the body by $1^{\circ} C$.
Heat capacity of a substance depends upon the material and mass of the body. Specific heat
capacity of a substance does not depend on the mass of the body. S.I. unit of heat capacity is $J K^{-1}$ and S.I. unit of specific heat capacity is $J kg ^{-1} K^{-1}$.
View full question & answer→