[5 Mark Question Answer]

Question 15 Marks

(i) (a) What is meant by specific heat capacity of a substance?
(b) Why does the heat supplied to a substance during its change of state not cause any rise in its temperature?
(ii) A student performs the following experiment in order to calculate specific latent heat of fusion of ice. He takes a calorimeter of mass 5.0 g containing 50 g of water at $30^{\circ} C$. Into this calorimeter he adds 5 g of dry ice at $0^{\circ} C$. When all the ice melts the final temperature recorded by him is $20^{\circ} C$ ?
(a) What is the total quantity of water in the calorimeter at $20^{\circ} C$ ?
(b) What quantities of heat are released by water and calorimeter in attaining the temperature of $20{ }^{\circ} C$.
(c) What is the total amount of heat gained by the ice?
(d) Calculate the value of latent heat of fusion of ice from the above calculations.
[Take SHC of water $=4.2 Jg ^{-10} C ^{-1}$ and SHC of copper $=0.4 Jg ^{-10} C ^{-1}$ ]
(iii) (a) Define heat capacity of a substance.
(b) Write the SI unit of heat capacity.
(c) What is the relationship between heat capacity and specific heat capacity of a substance?

Answer

(i) (a) The amount of heat energy required to raise the temperature of unit mass of a substance through 1 K is called its specific heat capacity.
(b) All the heat energy supplied during the change of state is utilised in increasing the intermolecular spaces, without any increase in the kinetic energy of molecules. Thus, as the average kinetic energy of molecules remain same, therefore, temperature does not rise.
(ii) (a) Total quantity of water in calorimeter at $20^{\circ} C =$ Initial mass of water + water formed from ice $=50 g+5 g= 5 5 g$
(b) Fall in the temperature of water and calorimeter $\left(\theta_{ F }\right)=(30-20)^{\circ} C =10^{\circ} C$
$\therefore$ Heat energy released by water $=m c \theta_{ F }=50 g \times 4.2 Jg ^{-19} C ^{-1} \times 10^{\circ} C =2100 J$
Heat energy released by calorimeter $=m c \theta_{ F }=5 g \times 0.4 Jg ^{-1} C ^{-1} \times 10^{\circ} C =20 J$
$\therefore$ Total heat energy released by water and calorimeter $=2100 J+20 J= 2 1 2 0 J$
(c) Rise in the temperature of water formed from ice $=(20-0)^{\circ} C =20^{\circ} C$
Heat energy gained by ice during melting $=m L_{ ice }=5 g \times L _{ ice }$
Heat energy gained by water at $0^{\circ} C =m c \theta_{ R }=5 g \times 4.2 Jg ^{-10} C ^{-1} \times 20{ }^{\circ} C =420 J$
Total heat energy gained by ice $=5 g \times L _{ lec }+ 4 2 0 J$
(d) By the principle of calorimetry :
Heat gained by ice $=$ Heat released by water and calorimeter
$\Rightarrow 5 g \times L _{ io }+420 J=2120 J$
$\Rightarrow \quad 5 g \times L _{i c e}=(2120-420) J =1700 J$
$\therefore$ Latent heat of fusion of ice $\left(L_{\text {ice }}\right)=\frac{1700 J}{5 g}=340 Jg ^{-1}$

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Question 25 Marks

(i) A copper vessel of mass 100 g contains 150 g of water at $50^{\circ} C$. How much ice is needed to cool it to $5^{\circ} C$ ?
Given : Specific heat capacity of copper $=0.4 J g ^{-1}{ }^{\circ} C ^{-1}$
Specific heat capacity of water $=4.2 Jg ^{-1}{ }^{\circ} C ^{-1}$
Specific latent heat of fusion of ice $=336 Jg ^{-1}$
(ii) A piece of ice of mass 60 g is dropped into 140 g of water at $50^{\circ} C$. Calculate the final temperature of water when all the ice has melted. (Assume no heat is lost to the surrounding)
Specific heat capacity of water $=4.2 Jg ^{-1} K^{-1}$.
Specific latent beat of fusion of ice $=336 Jg ^{-1}$.
(iii) (a) Define specific latent heat of fusion of ice.
(b) What happens to the heat supplied to the substance when the heat supplied causes no change in the temperature of the substance?

Answer

(i) Let mass of ice $=m$
$\therefore$ Heat gained by ice to form water at $0^{\circ} C =m L=m \times 336$
Heat gained by water at $0^{\circ} C$ to attain temperature of $5^{\circ} C =m c \theta_{ R }=m \times 4,2 \times 5=21 m$
$\therefore$ Total heat gained $=336 m+21 m=357 m$
Heat lost by vessel $=m c \theta_f=100 \times 0.4 \times 45=1800 J$
Heat lost by water $=m c \theta_f=150 \times 4.2 \times 45=28350 J$
$\therefore$ Total heat lost $-28350+1800=30150 J$
Heat gained $=$ Heat lost
$\Rightarrow 357 m=30150$
$\therefore m=\frac{30150}{357}= 8 4 . 4 5 g$.
$\begin{array}{l}\text { (ii) Let final temperature }=\theta \text {, then } \\ \text { Heat gained by the ice }=m L+m c \theta=60 \times 336+60 \times 4.2 \times \theta=20160+252 \theta \\ \text { Heat lost by water }=140 \times 4.2 \times(50-\theta)=588(50-\theta)=29400-588 \theta \\ \text { Now, Heat gained }=\text { Heat lost }\end{array}$
$\Rightarrow 20160+252 \theta=29400-588 \theta$
$\Rightarrow 840 \theta=9240$
$\therefore \theta=\frac{9240}{840}=11^{\circ} C$
Thus, final temperature $=11^{\circ} C$
(iii) (a) The amount of heat energy required to melt one kilogram of ice ( 1 g in case of CGS system) at its melting point, without any rise in termperature is called specific latent heat of fusion of ice
(b) The heat supplied to the substance is stored within the molecules in the form of potential energy. This stored heat energy only increases the intermolecular spaces and decreases the intermolecular forces..

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Question 35 Marks

(i) (a) Define calorie.
(b) What do you understand by the term latent heat of fusion?
(ii) A piece of metal of mass 50 g is at $100^{\circ} C$. It is placed on a block of ice. Calculate the amount of ice which melts, if the specific heat capacity of metal is $0.8 Jg ^{-1} C ^{-1}$ and specific latent heat of fusion of ice is $340 Jg ^{-1}$.
(iii) A vessel of negligible thermal capacity contains 60 g of water at $40^{\circ} C$. In this water is placed 10 g of ice at $-10^{\circ} C$. If all the ise melts calculate the final temperature of water.
Specifle heat capacity of ice $=2100 Jkg ^{-19} C ^{-1}$
Specific heat capacity of water $=4200 Jkg ^{-19} C ^{-1}$
Specific latent heat of fission of ice $=336 \times 10^3 J / kg$

Answer

(i) (a) Calorie: It is the amount of heat energy required to raise the temperature of 1 g of water through $1^{\circ} C$.
(b) The beat energy supplied to a solid so as to change it into the liquid state, without any rise in temperature is called latent heat of fusion. During fusion, the beat energy supplied to the solid is used in increasing the intermolecular space within the substance and is stored as potential energy of the molecules. The temperature of the substance does not rise until whole of solid is converted into liquid.
(ii) Heat gained by ice = Heat lost by metal
$\Rightarrow m L_{j a x}=m c \theta_F$
$\Rightarrow m \times 340 Jg ^{-1}=50 g \times 0.8 Jg ^{-1}{ }^{\circ} C ^{-1} \times 100^{\circ} C$
$\therefore m=\frac{50 \times 0.8 \times 100}{340} g=11.76 g$
(iii) Let the final temperature of water $=x$.
$\therefore$ Fall in temperature of water at $40^{\circ} C \left(B_\rho\right)=\left(40^{\circ} C -x\right)$
Rise in temperature of water formed from ice $\left(\theta_R\right)=(x-0)=x$
Heat gained by ice at $-10^{\circ} C$ to form ice at $0^{\circ} C =m a 0_{ R }-10 g \times 2.1 Jg ^{-1}{ }^{\circ} C ^{-1} \times 10^{\circ} C =210 J$
Heat gained by ice at $0^{\circ} C$ to melt $= mL _{\text {jes }}=10 g \times 336 Jg ^{-1}=3360 J$
Heat gained by water at $0^{\circ} C =m \theta_{ R }=10 g \times 4.2 Jg ^{-10} C ^{-1} \times x=42 x J ^{\circ} C ^{-1}$
$\therefore$ Total heat gained by ice $=210 J+3360 J+42 x J^{\circ} C ^{-1}=3570 J+42 x J^{\circ} C ^{-1}$
Heat lost by water at $40^{\circ} C =m c \theta_f=60 g \times 4.2 Jg ^{-1 \circ} C ^{-1} \times\left(40^{\circ} C -x\right)=10080 J-252 x J^{\circ} C ^{-1}$
Now, Heat gained = Heat lost
$\Rightarrow 3570 J+42 x J^{\circ} C ^{-1}=10080 J-252 \times J ^{\circ} C ^{-1}$
$\Rightarrow 42 \times J ^{\circ} C ^{-1}+252 \times J ^{\circ} C ^{-1}=10080 J-3570 J$
$294 x^{\circ} C ^{-1}=6510$
$x=\frac{6510}{294}^{\circ} C =22.14^{\circ} C$

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Question 45 Marks

(i) A piece of ice of mass 40 g is dropped into 200 g of water at $50^{\circ} C$. Calculate the final temperature of water after all the ice has melted. (specific heat capacity of water $=4200 J / kg { }^{\circ} C$, specific latent heat of fusion of ice $\left.=336 \times 10^3 H / kg \right)$
(ii) A piece of iron of mass 2.0 kg has a thermal capacity of $966 J / \circ C$.
(a) How mach heat is needed to warm it by $15^{\circ} C$ ?
(b) What is its specific heat capacity in SI units?
(iii) Explain the following :
(a) Why is water used as coolant in motor car radiators?
(b) Why it takes more time to change water at $100^{\circ} C$ to steam at $100^{\circ} C$ compared to heating same mass of water from $0^{\circ} C$ to $100^{\circ} C$ ?

Answer

(i) Let the final temperature $=x$
$\therefore$ Rise in temperature $\left(O_R\right)=(x-0)=x$; Fall in temperature $\left(\theta_d\right)=(S 0-x)$ Heat gained by ice to form water at $0^{\circ} C =m L_{\text {ice }}=\frac{40}{1000}(kg) \times 336 \times 10^3 \frac{J}{ kg }=13440 J$ Heat gained by $\frac{40}{1000}(kg)$ of water to attain temperature $x=m c \theta_{ R }=\frac{40}{1000} \times 4200 \times x=168 x$
$\therefore$ Total heat gained $-13440+168 x$ Heat lost by water at $50^{\circ} C =m c \theta_f=\frac{200}{1000}(kg) \times 4200 \frac{J}{ kg ^{\circ} C } \times(50-x)-840(50-x)-42000-840 x$ By the principle of calorimetry,
Heat gained $=$ Heat lost
$\Rightarrow 13440+168 x=42000-840 x$
$\Rightarrow 168 x+840 x=42000-13440$
$\Rightarrow 1008 x=28560 \Rightarrow x=\frac{28560}{1008}=28.33^{\circ} C$
(ii) Mass of iron $=2.0 kg$; Thermal capacity $=966 J /{ }^{\circ} C$.
(a) Heat needed to warm iron by $15^{\circ} C =$ Thermal capacity $\times$ Rise in temperature
$=966 \frac{J}{{ }^{\circ} C} \times 15^{\circ} C=14490 J$
(b) Specific heat capacity of iron $=\frac{\text { Thermal capacity of iron }}{\text { Mass of iron }}=\frac{966 J / C }{2.0 kg}=483 J kg ^{-1}{ }^{\circ} C ^{-1}$.
(iii) (a) Water has the highest specific heat capacity of $4200 kJ / kg ^{\circ} C$. Thus, water can absorb maximum amount of heat energy as compared to another substance for the same rise of temperature and hence is used as coolant in motor car radiators.
(b) Assume there is 1 kg of water at $0^{\circ} C$.
$\therefore$ Amount of heat energy required to bring it $100^{\circ} C =m c \theta_{ R }=1 kg \times 4200 \frac{J}{ kg } \times 100^{\circ} C =420000 J$
Also, amount of heat energy required to change water at $100^{\circ} C$ into steam
$-mL_{\text {tasm }}-1 kg \times 2268000 Jkg^{-1}-2268000 J$
From the above data it is clear that more heat energy is required to change the state of water to steam rather than to bring it to $100^{\circ} C$. Thus, it will take longer time to change water into steam as compared to bringing it to $100^{\circ} C$.

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Question 55 Marks

(i) 50 g of ice at $0^{\circ} C$ is added to 300 g of a liquid at $30^{\circ} C$. What will be the final temperature of the mixture when all the ice has melted? The specific heat capacity of the liquid is $2.65 Jg ^{-19} C ^{-1}$ while that of water is $4.2 Jg ^{-1}{ }^{\circ} C ^{-1}$. Specific latent heat of fision of ice $=336 Jg ^{-1}$.
(ii) A metal drill of power 500 W , drills a bole in a metal cube of mass 0.25 kg in 6.5 s . If the specific heat capacity of metal cube is $130 Jkg ^{-1}{ }^{\circ} C ^{-1}$, calculate
(a) Energy generated by the drill in ls.
(b) Energy generated by the drill in 6.5 s.
(c) If $t^{\circ} C$ is the rise in temperature, state the amount of heat energy absorbed by metal cube in terms of $t$.
(d) Write down an equation and calculate the value of t .
(iii) Explain the following
(a) Water is used in hot water bottles for fomentation purposes.
(b) A wise farmer always waters his fields in the evening, if there is a forecast for frost.
(c) The specific heat of fusion of lead is $27 Jg ^{-1}$. What do you understand from the statement?

Answer

(i) Let the final temperature of mixture is ' $x$ '.
$\therefore$ Rise in temperature $\theta_{ R }=(x-0)=x$.
Fall in temperature of liquid $\theta_f=(30-x)$.
Heat given out by liquid $=m c \theta_f=300 \times 2.65(30-x)=23850-795 x$
Heat absorbed by ice to form water at $0^{\circ} C =m L_{\text {iee }}=50 \times 336=16800$.
Heat absorbed by water at $0^{\circ} C$ to attain temperature $x=m c \theta_{ R }=50 \times 4.2 \times x=210 x$
$\therefore$ Total heat absorbed by ice $=16800+210 x$
Heat absorbed by ice $=$ Heat given out by liquid
$\Rightarrow 16800+210 x=23850-795 x$
$\Rightarrow 210 x+795 x=23850-16800$
$\Rightarrow 1005 x=7050$
$\therefore x =\frac{7050}{1005}=7.01^{\circ} C$
Final temperature of mixture $=7.01^{\circ} C$.
(ii) (a) Energy generated by drill in $1 s= P \times t =500 \frac{J}{ s } \times 1 s=500 J$.
(b) Energy generated by drill in $6.5 s=500 \frac{J}{ s } \times 6.5 s=3250 J$.
(c) Energy absorbed by metal cube $=0.25 \times 130 \times t=32.5 t$
(d) Energy absorbed by metal cube $=$ Energy generated by drill
$32.5 t =3250$
$\therefore t =\frac{3250}{32.5}=100^{\circ} C$
(iii) (a) Water does not cool quickly due to its large specific beat capacity. So a hot water bottle provides heat energy for fomentation for a longer period.
(b) When temperature falls below $0^{\circ} C$, water in the capillaries of plants would freeze and thus capillaries burst due to anomalous expansion of water on cooling. As a result plants would die and the crop would get destroyed. In order to save crops on such cold nights, farmers water their fields in the evening. As water has high specific heat capacity it liberates lot of heat energy as it cools and does not allow the temperature in the surrounding of plants to fall below $0^{\circ} C$.
(c) It means that 27 J of heat energy is required to melt 1 g of lead at its melting point, without any rise in temperature. or 27 J of energy is liberated during solidification of 1 g of lead at its melting point, without any change in temperature.

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Physics [5 Mark Question Answer]

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