1 Marks Question

Question 11 Mark

When a solid melts or a liquid boils, the temperature does not increase even when heat is supplied. Where does the energy go?

Answer

The energy is consumed in the process of changing state of the material so as from liquid to vapor or solid to liquid. the heat involved changes form of material and the temperature doesn't increases.

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Question 21 Mark

A block of mass 100g slides on a rough horizontal surface. If the speed of the block decreases from $10ms^{-1}$ to $5ms^{-1}$, find the thermal energy developed in the process.

Answer

Given,
Mass of the block = 100g = 0.1kg
Initial speed of the block = 10m/s
Final speed of the block = 5m/s
Initial kinetic energy of the block $=\frac{1}{2}\times0.1\times10^2=5\text{J}$
Final kinetic energy of the block $=\frac{1}{2}\times0.1\times5^2=1.25\text{J}$
Change in kinetic energy of the block = 5 - 1.25 = 3.75J
Thermal energy developed is equal to the change in kinetic energy of the block. Thus,
Thermal energy developed in the process = 3.75J

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Question 31 Mark

Four 2cm × 2cm × 2cm cubes of ice are taken out from a refrigerator and are put in 200ml of a drink at 10°C.

Find the temperature of the drink when thermal equilibrium is attained in it.
If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice $= 900kg-m^{-3}$, density of the drink $= 1000kg-m^{-3}$, specific heat capacity of the drink $= 4200Jkg^{-1}-K^{-1}$, latent heat of fusion of ice $= 3.4 × 10^5J-kg^{-1}$.

Answer

Given,

Number of ice cubes$ = 4$
Volume of each ice cube$= (2 × 2 × 2) = 8cm^3$
Density of ice $= 900kg\ m^{-3}$
Total mass of ice, $m_i= (4 × 8 ×10^{-6} × 900) = 288 × 10^{-4}\ kg$
Latent heat of fusion of ice, $L_i= 3.4 × 10^5\ J\ kg^{-1}$
Density of the drink $= 1000kg\ m^{-3}$
Volume of the drink $= 200ml$
Mass of the drink $= (200 × 10^{-6}) × 1000kg$
Let us first check the heat released when temperature of 200ml changes from $10^\circ C$ to $0^\circ C.$
$H_w= (200 × 10^{-6}) × 1000 × 4200 × (10 - 0) = 8400J$
Heat required to change four $8cm^3$ ice cubes into water $(H_i) = m_iL_i= (288 × 10^{-4}) × (3.4 × 0^5) = 9792J$
Since the heat required for melting the four cubes of the ice is greater than the heat released by water $( H_i > H_w)$, some ice will remain solid and there will be equilibrium between ice and water
Thus, the thermal equilibrium will be attained at $0^\circ C$

Equilibrium temperature of the cube and the drink $= 0°C$

Let M be the mass of melted ice
Heat released when temperature of 200ml changes from $10^\circ C$ to $0^\circ C$ is given by
$H_w= (200 × 10^{-6}) × 1000 × 4200 × (10 - 0) = 8400J$
Thus,
$M × (3.4 × 10^5) = 8400J$
Therefore,
$M = 0.0247Kg = 25g$

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Question 41 Mark

Should a thermometer bulb have large heat capacity or small heat capacity?

Answer

Thermometer bulb must have small heat capacity so as with very small amount of heat the temperature rises up to prominent amount.

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Question 51 Mark

A van of mass 1500kg travelling at a speed of $54kmh^{-1}$ is stopped in 10s. Assuming that all the mechanical energy lost appears as thermal energy in the brake mechanism, find the average rate of production of thermal energy is $cals^{-1}$.

Answer

Given,
Mass of van, m = 1500kg
Speed of van, v = 54km/h
$=54\times\Big(\frac{5}{18}\Big)=15\text{m/s}$
Total kinetic energy of the van is given by,
$\text{K}=\frac{1}{2}\text{mv}^2$
$\text{K}=\frac{1}{2}\times1500\times(15)^2$
$\text{K}=750\times225$
$\text{K}=168750\text{J}$
$\text{K}=\frac{168750}{4.2}$
$\text{K}=40178.57\ \text{cal}$
Loss in total energy of the van = 40178cal
Loss in energy per second $=\frac{40178}{10}=4017.8\approx4000\text{cal/sec}$
$\therefore$ Average rate of production of thermal energy $\approx4000\text{cal/sec}$

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Question 61 Mark

Calculate the time required to heat 20kg of water from 10°C to 35°C using an immersion heater rated 1000W. Assume that 80% of the power input is used to heat the water. Specific heat capacity of water $= 42000J\ kg^{-1}\ K^{-1}$.

Answer

Given,
Power rating of the immersion rod, $P = 1000W$
Specific heat of water, $S = 4200J\ kg^{-1}\ K^{-1}$
Mass of water,$M = 20kg$
Change in temperature, $\triangle\text{T}=25^\circ\text{C}$
Total heat required to raise the temperature of 20kg of water from 10°C to 35°C is given by
$\text{Q} = \text{M} \times\text{S}\times\triangle\text{T}$
$\text{Q} = 20 \times4200\times25$
$\text{Q} = 20 \times4200\times25$
$\text{Q} = 21\times10^5\text{J}$
Let the time taken to heat 20kg of water from 10°C to 35°C be t. Only 80% of the immersion rod's heat is useful for heating water. Thus,
Energy of the immersion rod utilised for heating the water $= t × (0.80) × 1000J$
$t × (0.80) × 1000J = 21 × 10^5\ J$
$ \text{t}=\frac{21\times10^5}{800}=2625\text{s}$
$\Rightarrow \text{t}=\frac{2625}{60}=43.75\text{min}\approx44\text{min}$

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Question 71 Mark

A brick weighing 4.0kg is dropped into a 1.0m deep river from a height of 2.0m. Assuming that 80% of the gravitational potential energy is finally converted into thermal energy, find this thermal energy is calorie.

Answer

Given,
Mass of the brick, m = 4kg
Total vertical distance travelled by the brick, h = 3m
Percentage of gravitational potential energy converted to thermal energy = 80
Total change in potential energy of the brick = mgh = 4 × 10 × 3 = 120J
Thermal Energy $=120\times\frac{80}{100}=96\text{J}$
Thermal energy in calories is given by,
$\text{U}=\frac{96}{4.2}=22.857\text{cal}\approx23\text{cal}$

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Question 81 Mark

1 kg of ice at $0^{\circ} C$ is mixed with 1 kg of steam at $100^{\circ} C$. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice $=3.36 \times 10^3 J kg ^{-1}$ and latent heat of vaporization of water $=2.26 \times 10^6 J kg ^{-1}$.

Answer

Given,
Amount of ice at $0^{\circ} C =1 kg$
Amount of steam at $100^{\circ} C =1 kg$
Latent heat of fusion of ice $=3.36 \times 10^3 J kg ^{-1}$
Latent heat of vapourisation of water $=2.26 \times 10^6 J kg ^{-1}$
We can observe that the latent heat of fusion of ice $\left(3.36 \times 10^5 Jg ^{-1}\right)$ is smaller that latent heat of vapouisation of water $\left(2.26 \times 10^6\right)$. Therefore, ice will first change into water as less heat is required for this and there will be equilibrium between steam and water.
Heat absorbed by the ice when it changes into water $\left(Q_1\right)=1 \times\left(3.36 \times 10^5\right) J$
Heat absorbed by the water formed to change its temperature from $0^{\circ} C$ to $100^{\circ} C \left( Q _2\right)=1 \times 4200 \times 100=4.2 \times 10^5$ J
Total heat absorbed by the ice to raise the temperature to $100^{\circ} C , Q = Q _1+ Q _2=3.36 \times 10^5+4.2 \times 10^5=(3.36+$ $4.2) \times 10^5=7.56 \times 10^5 J$
The heat required to change ice into water at $100^{\circ} C$ is supplied by the steam. This heat will be released by the steam and will then change into water.
If all the steam gets converted into water, heat released by steam, $Q ^{\prime}=1 \times\left(2.26 \times 10^6\right) J =2.26 \times 10^6 J$
Amount of heat released is more than that required by the ice to get converted into water at $100^{\circ} C$. Thus,
$\text { Extra heat }=Q-Q^{\prime}$
$=(2.26-0.756) \times 10^6$
$=1.506 \times 10^6$
Let the mass of steam that is condensed into water be m . Thus,
$m=\frac{7.56 \times 10^5}{2.26 \times 10^6}$
$=0.335 kg=335 gm$
Total amount of water at $100^{\circ} C =1000+335=1335 g=1.335 g$
Steam left $=1-0.335=0.665 kg=665 gm$

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Question 91 Mark

In Regnault's apparatua for measuring specific heat capacity of a solid, there is an inlet and an outlet in the steam chamber. The inlet is near the top and the outlet is near the bottom. Why is it better than the opposite choice where the inlet is near the bottom and the outlet is near the top?

Answer

Inlet at the top allows material to got down due to gravity and material stays more time in steam chamber.

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Question 101 Mark

On a winter day the temperature of the tap water is $20^{\circ} \mathrm{C}$ whereas the room temperature is $5^{\circ} \mathrm{C}$. Water is stored in a tank of capacity $0.5 \mathrm{~m}^3$ for household use. If it were possible to use the heat liberated by the water to lift a 10 kg mass vertically, how high can it be lifted as the water comes to the room temperature? Take $\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}$.

Answer

Given,
Initial temperature of the water, $\mathrm{T}_{\mathrm{i}}=20^{\circ} \mathrm{C}$
Final temperature of the water (room temperature), $\mathrm{T}_{\mathrm{f}}=5^{\circ} \mathrm{C}$
Change in temeprature, $\triangle \mathrm{T}=20^{\circ} \mathrm{C}-5^{\circ} \mathrm{C}=15^{\circ} \mathrm{C}$
Volume of water $=0.5 \mathrm{~m}^3$
Density of water, $d=1000 \mathrm{~kg} / \mathrm{m}^3$
Mass of the water, $\mathrm{M}=(0.5 \times 1000) \mathrm{kg}=500 \mathrm{~kg}$
Heat liberated as the temperature of water changes from $20^{\circ} \mathrm{C}$ to $5^{\circ} \mathrm{C}$ is given by,
$ \mathrm{Q}=\mathrm{M} \times \mathrm{S} \times \Delta \mathrm{T} $
$ \mathrm{Q}=(500 \times 4200 \times 15) \mathrm{J} $
$ \mathrm{Q}=(75 \times 420 \times 1000) \mathrm{J} $
$ \mathrm{Q}=31500 \times 1000 \mathrm{~J} $
$ \mathrm{Q}=315 \times 10^5 \mathrm{~J}$
Let the height to which the mass is lifted be $h$.
The energy required to lift the block $=\mathrm{mgh}=10 \times 10 \times \mathrm{h}=100 \mathrm{~h}$
Acording to the question,
$ Q=m g h $
$ \Rightarrow 100 \mathrm{~h}=315 \times 10^5 \mathrm{~J} $
$ \Rightarrow \mathrm{~h}=315 \times 10^3 \mathrm{~m}=315 \mathrm{~km}$

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Question 111 Mark

A bullet of mass 20g enters into a fixed wooden block with a speed of $40ms^{-1}$ and stops in it. Find the change in internal energy during the process.

Answer

Given,
Mass of bullet, $m = 20g = 0.02kg$
Initial velocity of the bullet, $u = 40ms^{-1}$
Final velocity of the bullet $= 0ms^{-1}$
Initial kinetic energy of the bullet $=\frac{1}{2}\text{mu}^2$
$=\frac{1}{2}\times0.02\times40\times40=16\text{J}$
Final kinetic energy of the bullet = 0
Change in energy of the bullet = 16J
It is given that the bullet enters the block and stops inside it. The total change in its kinetic energy is responsible for the change in the internal energy of the block.
$\therefore$ Change in internal energy of the block = Change in energy of the bullet = 16J.

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Question 121 Mark

Two blocks of masses 10kg and 20kg moving at speeds of $10ms^{-1}$ and $20ms^{-1}$ respectively in opposite directions, approach each other and collide. If the collision is completely inelastic, find the thermal energy developed in the process.

Answer

Given,
Mass of the first block, $\mathrm{m}_1=10 \mathrm{~kg}$
Mass of the second block, $\mathrm{m}_2=20 \mathrm{~kg}$
Initial velocity of the first block, $u_1=10 \mathrm{~m} / \mathrm{s}$
Initial velocity of the second block, $u_2=20 \mathrm{~m} / \mathrm{s}$
Let the velocity of the blocks after collision be v .
Applying conservation of momentum, we get
$ m_2 u_2-m_1 u_1=\left(m_1+m_2\right) v $
$ \Rightarrow 20 \times 20-10 \times 10=(10+20) v $
$ \Rightarrow 400-100=30 \mathrm{v} $
$ \Rightarrow 300=30 \mathrm{v} $
$ \Rightarrow v=10 \mathrm{~m} / \mathrm{s}$

Initial kinetic energy is given by,
$\text{K}_\text{i}=\frac{1}{2}\text{m}_1\text{u}_1^2+\frac{1}{2}\text{m}_2\text{u}_2^2$
$\text{K}_\text{i}=\frac{1}{2}\times10\times(10)^2+\frac{1}{2}\times20\times(20)^2$
$\text{K}_\text{i}=500+4000=4500$
Final kinetic energy is given by,
$\text{K}_\text{f}=\frac{1}{2}(\text{m}_1+\text{m}_2)\text{v}^2$
$\text{K}_\text{f}=\frac{1}{2}(10+20)(10)^2$
$\text{K}_\text{f}=\Big(\frac{30}{2}\Big)\times100=1500$
$\therefore$ Total change in KE = 4500J - 1500J = 3000J
Thermal energy developed in the process = 3000J

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Question 131 Mark

A cube of iron (density $=8000 \mathrm{~kg} \mathrm{~m}^{-3}$, specific heat capacity $=470 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ ) is heated to a high temperature and is placed on a large block of ice at $0^{\circ} \mathrm{C}$. The cube melts the ice below it, displaces the water and sinks. In the final equilibrium position, its upper surface just goes inside the ice. Calculate the initial temperature of the cube. Neglect any loss of heat outside the ice and the cube. The density of ice $=900 \mathrm{~kg} \mathrm{~m}^{-3}$ and the latent heat of fusion of ice $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$.

Answer

Given,
Density of the iron cube $=8000 \mathrm{~kg} \mathrm{~m}^{-3}$
Density of the ice cube $=900 \mathrm{~kg} \mathrm{~m}^{-3}$
Specific heat capacity, $\mathrm{S}=470 \mathrm{Jg}^{-1} \mathrm{~K}^{-1}$
Latent heat of fusion of ice, $L=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$
Let the volume of the cube be $V$
Volume of water displaced $=\mathrm{V}$
Mass of cube, $\mathrm{m}=8000 \mathrm{~V} \mathrm{~kg}$
Mass of the ice melted, $\mathrm{M}=900 \mathrm{~V}$
Let the initial temperature of the iron cube be TK. Then, Heat gained by the ice $=$ Heat lost by the iron cube
$ \Rightarrow \mathrm{m} \times \mathrm{S} \times(\mathrm{T}-273)=\mathrm{M} \times \mathrm{L} $
$ \Rightarrow 8000 \mathrm{~V} \times 470 \times(\mathrm{T}-273)=900 \mathrm{~V} \times\left(3.36 \times 10^5\right) $
$ \Rightarrow 376 \times 10^4 \times(\mathrm{T}-273)=3024 \times 10^5$
$\Rightarrow\text{T}=\frac{30240+102648}{376}$
$\Rightarrow\text{T}=\frac{132888}{376}\text{K}$
$\Rightarrow\text{T}=353.425\text{K}\approx353\text{K}$
$\Rightarrow\text{T}=353\text{K}-273\text{K}$
$\Rightarrow\text{T}=80^\circ\text{C}$

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Question 141 Mark

In a calorimeter, the heat given by the hot object is assumed to be equal to the heat taken by the cold object. Does it mean that heat of the two objects taken together remaina constant?

Answer

Heat is measure of transfer of energy there is no heat in the body or heat of the body. So heat of the two objects doesn't make any sense.

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Question 151 Mark

A 50kg man is running at a speed of $18kmh^{-1}$. If all the kinetic energy of the man can be used to increase the temperature of water from 20°C to 30°C, how much water can be heated with this energy?

Answer

Given,
Mass of the man, m = 50kg
Speed of the man, $\text{v}=18\text{km/h}$
$=18\times\frac{5}{18}=5\text{m/s}$
Kinetic energy of the man is given by,
$\text{K}=\frac{1}{2}\text{mV}^2$
$\text{K}=\Big(\frac{1}{2}\Big)50\times5^2$
$\text{K}=25\times25=625\text{J}$
Specific heat of the water, s = 4200J/Kg-K
Let the mass of the water heated be M.
The amount of heat required to raise the temperature of water from 20°C to 30°C is given by,
$\text{Q}=\text{ms}\triangle\text{T}=\text{M}\times4200\times(30-20)$
$\text{Q}=42000\text{M}$
According to the question,
$\text{Q}=\text{K}$
$\Rightarrow42000\text{M}=625 $
$\Rightarrow\text{M}=\frac{625}{42}\times10^{-3}$
$\Rightarrow\text{M}=14.88\times10^{-3}$
$\Rightarrow\text{M}=15\text{g}$

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Question 161 Mark

The calorie ia defined as 1cal = 4.186 joule. Why not as 1cal = 4J to make the conversions easy?

Answer

As cal is very small unit energy may be calculated in Kcal so in this case difference between 4.186 and 4 becomes large.

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Question 171 Mark

A ball is dropped on a floor from a height of 2.0m. After the collision it rises up to a height of 1.5m. Assume that 40% of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is $800JK^{-1}$.

Answer

Height of the floor from which ball is dropped, $h_1= 2.0m$
Height to which the ball rises after collision, $h_2= 1.5m4$
Let the mass of ball be m kg.
Let the speed of the ball when it falls from $h_1$ and $h_2$ be $v_1$ and $v_2$, respectively.
$\text{v}_1=\sqrt{2\text{gh}_1}=\sqrt{2\times10\times2}=\sqrt{40}\text{m/s}$
$\text{v}_2=\sqrt{2\text{gh}_2}=\sqrt{2\times10\times1.5}=\sqrt{30}\text{m/s}$
Change in kinetic energy is given by,
$\triangle\text{K}=\frac{1}{2}\times\text{m}\times40-\Big(\frac{1}{2}\text{m}\Big)\times30=\Big(\frac{10}{2}\Big)\text{m}$
$\Rightarrow\triangle\text{K}=5\text{m}$
If the position of the ball is considered just before hitting the ground and after its first collision, then 40% of the change in its KE will give the change in thermal energy of the ball. At these positions, the PE of the ball is same. Thus,
Loss in PE = 0
The change in kinetic energy is utilised in increasing the temperature of the ball.
Let the change in temperature be $\triangle\text{T}.$ Then,
$\Big(\frac{40}{100}\Big)\times\triangle\text{K}=\text{m}\times800\times\triangle\text{T}$
$\Big(\frac{40}{100}\Big)\times\frac{10}{2}\text{m}=\text{m}\times800\times\triangle\text{T}$
$\Rightarrow\triangle\text{T}=\frac{1}{400}=0.0025$
$\Rightarrow2.5\times10^{-3}{^\circ}\text{C}$

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Question 181 Mark

A copper cube of mass 200g slides down on a rough inclined plane of inclination 37° at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in the temperature of the block as it slides down through 60cm. Specific heat capacity of copper $= 420Jkg^{-1}K^{-1}$.

Answer

Mass of copper cube, m = 200g = 0.2kg
Length through which the block has slided, l = 60cm = 0.6m
Since the block is moving with constant velocity, the net force on it is zero. Thus,
Force of friction, f = mg
Also, since the object is moving with a constant velocity, change in its K.E will be zero. As the object slides down, its PE decreases at the cost of increase in thermal energy of copper.
The loss in mchanical energy of the copper block = Work done by the frictional force on the copper block to a distanceof 60cm.
$\text{W}=\frac{\text{mg}}{\sin\theta}$
$\text{W}=0.2\times10\times0.6\sin37^\circ$
$\text{W}=1.2\times\Big(\frac{3}{5}\Big)=0.72$
Let the change in temperature of the block be $\triangle\text{T.}$
Thermal energy gained by block $=\text{ms}\triangle\text{T}=0.2\times420\times\triangle\text{T}=84\triangle\text{T}$
But $84\triangle\text{T}=0.72$
$\Rightarrow\triangle\text{T}=\frac{0.72}{84}=0.00857$
$\Rightarrow\triangle\text{T}=0.0086=8.6\times10^{-3}{^\circ}\text{C}$

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Question 191 Mark

Is heat a conserved quantity?

Answer

Heat is not conserved quantity as heat only exists when there is some energy transferred.

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Question 201 Mark

A metal block of density $6000 \mathrm{~kg} \mathrm{~m}^{-3}$ and mass 1.2 kg is suspended through a spring of spring constant $200 \mathrm{~N} \mathrm{~m}^{-1}$. The spring-block system is dipped in water kept in a vessel. The water has a mass of 260 g and the block is at a height 40 cm above the bottom of the vessel. If the support to the spring is broken, what will be the rise in the temperature of the water. Specific heat capacity of the block is $250 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ and that of water is $4200 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$. Heat capacities of the vessel and the spring are negligible.

Answer

Given,
Density of metal block, $d = 600kgm^{-3}$
Mass of metal block, $m = 1.2kg$
Spring constant of the spring, $k = 200Nm^{-1}$
Volume of the block, $\text{V}=\frac{1.2}{6000}=2\times10^{-4}\text{m}^3$
When the mass is dipped in water, it experiences a buoyant force and in the spring there is potential energy stored in it.
If the net force on the block is zero before breaking of the support of the spring, then kx + Vρg = mg
$200x + (2 × 10^{-4}) × (1000) × (10) = 12$
$\Rightarrow\text{x}=\frac{(12-2)}{200}$
$\Rightarrow\text{x}=\frac{10}{200}=0.05\text{m}$
The mechanical energy of the block is transferred to both block and water. Let the rise in temperature of the block and the water be $\triangle\text{T}.$
Applying conservation of energy, we get
$\frac{1}{2}\text{kx}^2+\text{mgh}-\text{V}\rho\text{gh}=\text{m}_1\text{s}_1\triangle\text{T}+\text{m}_2\text{s}_2\triangle\text{T}$
$\Rightarrow\frac{1}{2}\times200\times0.0025+1.2\times10\times\Big(\frac{40}{100}\Big)\\-2\times10^{-4}\times1000\times10\times\Big(\frac{40}{100}\Big)$
$\Rightarrow\Big(\frac{260}{1000}\Big)\times4200\times\triangle\text{T}+1.2\times250\times\triangle\text{T}$
$\Rightarrow0.25+4.8-0.8=1092\triangle\text{T}+300\triangle\text{T}$
$\Rightarrow1392\triangle\text{T}=4.25$
$\Rightarrow\triangle\text{T}=\frac{4.25}{1392}=0.0030531$
$\Rightarrow\triangle\text{T}=3\times10^{-3}{^\circ}\text{C}$

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Physics 1 Marks Question

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