**1.** 1. An aluminium vessel of mass $0\cdot5\ kg$ contains $0\cdot2\ kg$ of water at $20°C$. A block of iron of mass $0\cdot2\ kg$ at $100°C$ is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are $910\ J/kg-K,\ 470\ J/kg-K$ and $4200\ J/kg-K$ respectively.

Mass of aluminium = $0.5kg$, $\\$ Mass of Iron= $0.2kg$ $\\$ Sp heat of Iron=$100^o C = 373^o k$. $\\$ Sp heat of iron=$\frac{470J}{kg}-k$ $\\$ Heat again = $0.5 \times 910 (T-293)+0.2 \times 4200 \times (343 -T)$ $\\$ $=(T-292)(0.5 \times 910+0.2 \times 4200)$ $\\$ $\therefore$ Heat gain = Heat lost $\\$ Mass of water = $0.2 kg$ $\\$ Temp. of aluminium and water = $20^o C = 295^o k$ Sp heat of aluminium =$\frac{910J}{kg-k}$ $\\$ Sp of heat=$\frac{4200J}{kg-k}$ $\\$ Heat lost= $0.2 \times 470 \times (373-T$) $\\$ $\Rightarrow (T-292)(0.5 \times 910 + 0.2 \times 4200) = 0.2 \times 470 \times(373-T)$ $\\$ $\Rightarrow (T-292)(455+8400) = 49(373-T)$ $\\$ $\Rightarrow(T-293) \bigg(\frac{1295}{94}\bigg)$ = $(373-T)$ $\\$ $\Rightarrow(T-293) \times 14 = 373 -T$ $\\$ $\Rightarrow T= \frac{4475}{15} = 298k$ $\\$ $\therefore T = 298 - 273 = 25^o C$. $\\$ The final temp = $25^o C$.

**2.** 2.A piece of iron of mass $100 g$ is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent $10 g$ containing $240 g$ of water at $20°C$. The mixture attains an equilibrium temperature of $60°C$. Find the temperature of the furnace. Specific heat capacity of iron = $\frac{470J}{kg}-°C$.

mass of Iron=$100 g$ $\\$ mass of water=$240g$ $\\$ $S_{iron}=\frac{470J}{kg^o C}$ $\\$ The final temp=$25^ oC$. $\\$ water Eq of caloriemeter = $10g$ $\\$ Let the Temp of surface = $0^o C$ $\\$ Total heat gained = Total heat lost $\\$ So , $\frac{100}{1000} \times 470 \times (\theta -60) = \frac{250}{1000} \times 4200 \times (60-20)$ $\\$ $\Rightarrow 47\theta-47 \times 60 = 25\times 42 \times 40 $ $\\$ $\Rightarrow \theta=4200 + \frac{2820}{47}=\frac{44820}{47}= 953.61^oC $ $\\$

**3.** 3. The temperatures of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C, and when B and C are mixed, it is 23°C. What will be the temperature when A and C are mixed ?

The temp. of A = $12^o C$ $\\$ The temp. of B=$28^oC $ $\\$ The temp. of $\Rightarrow B+C=23^o$ $\\$ The temp. of B=$19^o C$ $\\$ The temp of $\Rightarrow A+B=16^o $ $\\$ In accordance with the principle of caloriemetry when $A$ & $B$ are mixed $M_{CA}(16-12)=M_{CB}(19-16) \Rightarrow CA4=CB3 \Rightarrow CA=\frac{3}{4}CB .....(1)$ $\\$ And when $B$ & $C$ are mixed $\\$ $M_{CB}(23-19)=M_{CC}(28-23)\Rightarrow 4CB=5CC\Rightarrow CC=\frac{4}{5}CB.....(2)$ $\\$ When $A$ and $c$ are mixed, if $T$ is the common temperature of mixture $\\$ $M_{CA}(T-12)=M_{CC}(28-T)$ $\\$ $\Rightarrow \bigg(\frac{3}{4}\bigg) CB(T-12)=\bigg(\frac{4}{5}\bigg)CB(28-T)$ $\\$ $\Rightarrow 15T-180=448-16T$ $\\$ $\Rightarrow T=\frac{628}{31}=20.258^o C$ = $20.3^oC$ $\\$

**4.** 4. Four $2 cm \times 2 cm \times 2 cm$ cubes of ice are taken out from a refrigerator and are put in $200 ml$ of a drink at $10°C$. $(a)$ Find the temperature of the drink when thermal equilibrium is attained in it. $(b)$ 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 = $\frac{900kh}{m^3}$, density of the drink =$\frac{1000kg}{m^3}$, specific heat capacity of the drink =$\frac{4200K}{kg-K} $, latent heat of
fusion of ice $- 3-4 \times \frac{10J}{kg}$.

$a)$Heat released when temperature of $200ml$ changes to $0^C$ $\\$ $(Heat_{water})$ = $200 \times 10^{-6} \times 1000 \times 4200 \times 10= 8400J$ $\\$ Heat required to change $ 4 \times 8 cm^3$ into ice will be left un melted and there will be equilibrium between ice and water and of course equilibrium temperature will be $0^C$ $\\$ $b)$ mass of ice melted = $m$ $m \times 3.4 \times 10^5 = 8400$ $\\$ $m=0.0247 Kg = 25Kg$

**5.** 5. Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed for evaporation is taken from the water- itself and the water is cooled down. Assume that a pitcher contains $10 kg$ of water and $0 2 g$ of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by $5°C$. Specific heat capacity of water = $\frac{4200J}{kg-^oC}$ and latent heat of vaporization of water = $2"27 \times \frac{10^s J}{kg}$

Total heat released when temperature drops by $5^oC(Q)=mc\theta=10 \times 4200 \times 5 = 210000 J$ $\\$ Rate of heat taken away when water evaporates = $0.002 \times 27 \times 10^6 =\frac{454J}{s}$ $\\$ Time = $\frac{Q}{454} =\frac{210000}{454} = 462.555 seconds = 7.70 min$ $\\$

**6.** A cube of iron $(density =\frac{8000kg}{m^3}$, specific heat capacity =$\frac{470J}{kg-K})$ is heated to a high temperature and is placed on a large block of ice at $0°C$. 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 =$\frac{900kg}{m^3}$ and the latent heat of fusion of ice =$336 \times \frac{10^5J}{kg}$

Let initial temperature = $T$ $\\$ Let volume of the cube = $V m^3$ $\\$ Mass of the cube$(m_c)$ = $8000V Kg$ $\\$ Volume of ice melted = $V$ $\\$ Mass of ice melted $(m_w)$ = $900V Kg$ $\\$ Heat liberated by cube = $m_c \times C_c \times (T-0) = 8000V \times 470 \times T = 3760000VT$ $\\$ Heat taken by the ice to melt = $900V \times 3.36 \times 10^5 = 302400000VJ$ $\\$ Equating both heats, we get $T = 80.42^oC$ $\\$

**7.** $1 kg$ of ice at $0°C$ is mixed with $1 kg$ of steam at $100°C$. What will be the composition of the system when the thermal equilibrium is reached ? Latent heat of fusion of ice = $3.36 \times \frac{10J}{kg}$ and latent heat of vaporization of
water = $2-26 \times \frac{10J}{kg}$ $\\$

We can get the latent heat of fusion is smaller than latent heat of vaporization. So, ice will change into water first because less heat is required for this and most popularly there will be equilibrium between stream and water. $\\$ Heat released when ice changed $(H_1) = 1 \times 3.36 \times 10^5 J$ $\\$ Heat when temperature of water changes from $0^oC$ to $100^oC (H_2) = 1 \times 4200 \times 100 = 420000 J$ $\\$ Total heat taken by the ice to change into water at $100^oC = H_1+ H_2 = 420000+336000 = 756000 J $ $\\$ So this heat will be released by the steam to change into water $\\$ Mass of steam changed into water = $m$$\\$ Heat = $m \times$ latent heat = $756000 $ $\\$ $m = 756000 \times 2.26 \times 10^6 = .3345$ $\\$ Total mass of water = $1+ .3345 = 1.3345 kg $ $\\$ Mass of steam = $1-0.3345 = 0.66548 Kg $ $\\$

**8.** Calculate the time required to heat$ 20 kg$ of water from $10°C$ to $35°C$ using an immersion heater rated $1000 W$. Assume that$ 80%$ of the power input is used to heat the water. Specific heat capacity of water =$\frac{4200J}{kg}$ $\\$

Power input = $80 % of 1000 = .8 \times 1000 = 800 W $ $\\$ Heat taken by water = $20 \times 2400 \times (35-10) = 2100000 $ $\\$ Time =$\frac{heat}{power} =\frac{2100000}{800} = 2625 seconds = 43.75 min$

**9.** On a winter day the temperature of the tap water is $20°C$ whereas the room temperature is $5°C$. Water is stored in a tank of capacity $0.5m^3$ 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 $g =\frac{10m}{s^2}$ $\\$

Volume of water = $ .5m $ $\\$ Density = $\frac{1000kg}{m^3}$ $\\$ heat liberated by water = $mass \times specific$ $heat$ $\times \theta = 1000 \times .5 \times 4200 \times (20-5) = 31500000 J $ $\\$ $Heat=mgh = 10 \times 10 \times h = 31500000 = 315000 m = 315 km $ $\\$

**10.** A bullet of mass $20 g$ enters into a fixed wooden block with a speed of $\frac{40m}{s}$ and stops in it. Find the change in internal energy during the process.

Mass of bullet = $20g = 0.02 g$ $\\$ Velocity of bullet = $\frac{40m}{s}$ $\\$ Total energy of bullet = $\frac{1}{2}mv^2$ $\\$ = $\frac{1}{2}(0.02)(40)^2$ $\\$ $16J$

**11.** A $50 kg$man is running at a speed of $\frac{18km}{h}$. 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 ?

Mass of the man = $50 kg$ $\\$ Velocity = $ \frac{18km}{hr} = \frac{5m}{s}$ $\\$ Change in temperature of water $(\theta) = 10^oC $ $\\$ Total energy = $\frac{1}{2}mv^2 $ $\\$ = $\frac{1}{2}(50)(5)^2 = 625J $ $\\$ We know , $Q = mc\theta $ $\\$ $625 = m \times 4200 \times 10 $ $\\$ $m = 0.01488 kg $ $\\$ $= 14.88 gm$

**12.** A brick weighing $40 kg$ 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 in calorie?

12 None

SolutionsMass = $4.0 kg$ $\\$ Height = $1m + 2m = 3m $ $\\$ Potential energy = $mgh$ $\\$ $4 \times 9.81 \times 3 = 117.72 J $ $\\$ Thermal energy produced = $117.72 \times 80%$ = $94.176 = \frac{94.176J}{4.187} cal = 22.492 cal $ $\\$

**13.** A van of mass $1500 kg$ travelling at a speed of $\frac{54km}{hr}$ is stopped in $10 s$. Assuming that all the mechanical energy lost appears as thermal energy in the brake mechanism, find the average rate of production of thermal energy in $\frac{}{}$

Mass = $1500 kg$ $\\$ Velocity= $\frac{54km}{hr} = \frac{15m}{sec}$ $\\$ Time to stop = $10 sec$ $\\$ Total energy of the system = $\frac{1}{2}(1500)(15)^2=168750 J= 40303.3198 cal $ $\\$ Energy loss in 10 sec = $40303.3198 cal$ $\\$ Rate of loss of energy =$\frac{4030.33198cal}{sec}$

**14.** A block of mass $100 g$ slides on a rough horizontal surface. If the speed of the block decreases from $\frac{10m}{s}$ to $\frac{5m}{s}$, find the thermal energy developed in the process.

Mass = $100g = 0.1 kg$ $\\$ Change in energy = $\frac{1}{2}mv_1^2-\frac{1}{2}mv_1^2$ $\\$ = $\frac{1}{2}(.1)(10^2-5^2) = 3.75 J$ $\\$

**15.** Two blocks of masses $10 kg$ and $20 kg$ moving at speeds of $\frac{10m}{s}$ and $\frac{20m}{s}$ respectively in opposite directions, approach each other and collide. If the collision is completely inelastic, find the thermal energy developed in the process.

Energy of 1st block = $\frac{1}{2} \times 10 \times 10^2 = 500J $ $\\$ Energy of 2nd block = $\frac{1}{2} \times 20 \times 20^2 = 4000 J $ $\\$

**16.** A metal block of density $\frac{6000kg}{m^3}$ and mass $1.2 kg$ is suspended through a spring of spring constant $\frac{200N}{m}$. 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 \frac{250J}{kg-K}$ and that of water is $|frac{4200J}{kg-K}$. Heat capacities of the vessel and the spring are negligible.

When spring is broken, block falls from $40 cm$ $\\$ Loss in energy = $mgh$ = $1.2 \times 9.8 \times 4 = 4.707J $ $\\$ This heat will rise temperature of block as well as water and also temperature of both will be same because they are in equilibrium. $\\$ $m_{block} \times C_{block} \times \theta +m_{water} \times C_{water}\times \theta =4.707$ $\\$ $\theta (0.26 \times 4200 + 1.2 \times 250) =4.707$ $\\$ $\theta = 3.38 \times$ $10^{-3}$ ${^o}C$ $\\$

**17.** A copper cube of mass $200 g$ 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 $60 cm$. Specific heat capacity of copper =$\frac{420J}{Kg-k}$.

Mechanical energy will be lost by friction only.$\\$ Here friction force $(f)$ = $mg$ $sin\theta$ = $0.2 \times 9.8sin$ $37$ = $1.17955 N $ $\\$ Work done (energy lost or thermal energy) $\\$ $(H)$ = $1.17955 \times 0.6 = 0.7077344 J $ $\\$ $H = mc\theta = 0.2 \times 420 \times \theta = .7077344 $ $\\$ $\theta$ = $8.4$ x $10^{-6}$ $^{o} C $ $\\$

**18.** A ball is dropped on a floor from a height of $2.0 m$. After the collision it rises up to a height of $1.5 m$. 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 $\frac{800J}{K}$

Initial potential energy = $mgh_1 = 9.8 \times 2m = 19.8 m J$ $\\$ Final potential energy = $9.8 \times 1.5m = 14.7m J $ $\\$ Mechanical loss =$19.8 - 14.7 = 5.1 J $ $\\$ $40 %$ of mechanical loss =.$4 \times 5.1 = 2.04 J $