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?
AnswerThe 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.
View full question & answer→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.
AnswerGiven,
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
View full question & answer→Question 31 Mark
Four $2 \mathrm{~cm} \times 2 \mathrm{~cm} \times 2 \mathrm{~cm}$ cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at $10^{\circ} \mathrm{C}$.
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 $=900 \mathrm{~kg}-\mathrm{m}^{-3}$, density of the drink $=1000 \mathrm{~kg}-\mathrm{m}^{-3}$, specific heat capacity of the drink $=4200 \mathrm{Jkg}^{-1}-\mathrm{K}^{-1}$, latent heat of fusion of ice $=3.4 \times 10^5 \mathrm{~J}-\mathrm{kg}^{-}$ 1.
Answera. Given,Number of ice cubes $=4$
Volume of each ice cube $=(2 \times 2 \times 2)=8 \mathrm{~cm}^3$
Density of ice $=900 \mathrm{~kg} \mathrm{~m}^{-3}$
Total mass of ice, $\mathrm{m}_{\mathrm{i}}=\left(4 \times 8 \times 10^{-6} \times 900\right)=288 \times 10^{-4} \mathrm{~kg}$
Latent heat of fusion of ice, $\mathrm{L}_{\mathrm{i}}=3.4 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$
Density of the drink $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$
Volume of the drink $=200 \mathrm{ml}$
Mass of the drink $=\left(200 \times 10^{-6}\right) \times 1000 \mathrm{~kg}$
Let us first check the heat released when temperature of 200 ml changes from $10^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$.
$H_w=\left(200 \times 10^{-6}\right) \times 1000 \times 4200 \times(10-0)=8400 J$
Heat required to change four $8 \mathrm{~cm}^3$ ice cubes into water $\left(\mathrm{H}_{\mathrm{i}}\right)=\mathrm{m}_{\mathrm{i}} \mathrm{L}_{\mathrm{i}}=\left(288 \times 10^{-4}\right) \times\left(3.4 \times 0^5\right)=9792 \mathrm{~J}$
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 $0^{\circ} \mathrm{C}$.
b. Equilibrium temperature of the cube and the drink $=0^{\circ} \mathrm{C}$
Let $M$ be the mass of melted ice.
Heat released when temperature of 200 ml changes from $10^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$ is given by,
$H_w=\left(200 \times 10^{-6}\right) \times 1000 \times 4200 \times(10-0)=8400 J$
Thus,
$\mathrm{M} \times\left(3.4 \times 10^5\right)=8400 \mathrm{~J}$
Therefore,
$\mathrm{M}=0.0247 \mathrm{Kg}=25 \mathrm{~g}$
View full question & answer→Question 41 Mark
Should a thermometer bulb have large heat capacity or small heat capacity?
AnswerThermometer bulb must have small heat capacity so as with very small amount of heat the temperature rises up to prominent amount.
View full question & answer→Question 51 Mark
A van of mass 1500 kg travelling at a speed of $54 \mathrm{kmh}^{-1}$ 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 is $\mathrm{cals}^{-1}$.
AnswerGiven, 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}$
View full question & answer→Question 61 Mark
Calculate the time required to heat 20 kg of water from $10^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{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 $=42000 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$.
AnswerGiven, Power rating of the immersion rod, $\mathrm{P}=1000 \mathrm{~W}$ Specific heat of water, $\mathrm{S}=4200 \mathrm{~J}{\mathrm{kg}^{-1}} \mathrm{~K}^{-1}$ Mass of water, $\mathrm{M}=$ 20 kg Change in temperature, $\triangle \mathrm{T}=25^{\circ} \mathrm{C}$ Total heat required to raise the temperature of 20 kg of water from $10^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$ is given by $\mathrm{Q}=\mathrm{M} \times \mathrm{S} \times \triangle \mathrm{T}$
$\mathrm{Q}=20 \times 4200 \times 25$
$\mathrm{Q}=20 \times 4200 \times 25$
$\mathrm{Q}=21 \times 10^5 \mathrm{~J}$
Let the time taken to heat 20 kg of water from $10^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{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 $=\mathrm{t} \times(0.80) \times 1000 \mathrm{~J} \times(0.80) \times$
$1000 \mathrm{~J}=21 \times 10^5 \mathrm{~J}=\frac{21 \times 10^5}{800}=2625 \mathrm{~s}$
$\Rightarrow \mathrm{t}=\frac{2625}{60}=43.75 \mathrm{~min} \approx 44 \mathrm{~min}$
View full question & answer→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.
AnswerGiven, 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}$
View full question & answer→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 Jkg ^{-1}$.
AnswerGiven, 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 ( $3.36 \times$ $\left.10^5 J kg ^{-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, 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$
View full question & answer→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?
AnswerInlet at the top allows material to got down due to gravity and material stays more time in steam chamber.
View full question & answer→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}$.
AnswerGiven, Initial temperature of the water, $T_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, $\mathrm{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 \triangle \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, $\mathrm{Q}=\mathrm{mgh} \Rightarrow 100 \mathrm{~h}=315 \times 10^5 \mathrm{~J} \Rightarrow \mathrm{~h}=315 \times 10^3 \mathrm{~m}=315 \mathrm{~km}$
View full question & answer→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.
AnswerGiven, 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.
View full question & answer→Question 121 Mark
Two blocks of masses 10 kg and 20 kg moving at speeds of $10 \mathrm{~ms}^{-1}$ and $20 \mathrm{~ms}^{-1}$ respectively in opposite directions, approach each other and collide. If the collision is completely inelastic, find the thermal energy developed in the process.
AnswerGiven, 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, $\mathrm{u}_1=$ $10 \mathrm{~m} / \mathrm{s}$ Initial velocity of the second block, $\mathrm{u}_2=20 \mathrm{~m} / \mathrm{s}$ Let the velocity of the blocks after collision be v . Applying conservation of momentum, we get $\mathrm{m}_2 \mathrm{u}_2-\mathrm{m}_1 \mathrm{u}_1=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{v} \Rightarrow 20 \times 20-10 \times 10=(10+20) \mathrm{v} \Rightarrow 400-100=30 \mathrm{v}$ $\Rightarrow 300=30 \mathrm{v} \Rightarrow \mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ Initial kinetic energy is given by,
$\mathrm{K}_{\mathrm{i}}=\frac{1}{2} \mathrm{~m}_1 \mathrm{u}_1^2+\frac{1}{2} \mathrm{~m}_2 \mathrm{u}_2^2$
$\mathrm{K}_{\mathrm{i}}=\frac{1}{2} \times 10 \times(10)^2+\frac{1}{2} \times 20 \times(20)^2$
$\mathrm{~K}_{\mathrm{i}}=500+4000=4500$
Final kinetic energy is given by, $\mathrm{K}_{\mathrm{f}}=\frac{1}{2}\left(\mathrm{~m}_1+\mathrm{m}_2\right) \mathrm{v}^2$
$\mathrm{K}_{\mathrm{f}}=\frac{1}{2}(10+20)(10)^2$
$\mathrm{~K}_{\mathrm{f}}=\left(\frac{30}{2}\right) \times 100=1500$
$\therefore$ Total change in $\mathrm{KE}=4500 \mathrm{~J}-1500 \mathrm{~J}=3000 \mathrm{~J}$
Thermal energy developed in the process $=3000 \mathrm{~J}$
View full question & answer→Question 131 Mark
A cube of iron (density $=8000 \mathrm{~kg} \mathrm{~m}^{-3}$, specific heat capacity $=470 \mathrm{Jg}^{-1} \mathrm{~K}^{-1}$ ) 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 = $3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$.
AnswerGiven, 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, $\mathrm{S}=470 \mathrm{~J}$ $\mathrm{kg}^{-1} \mathrm{~K}^{-1}$ Latent heat of fusion of ice, $\mathrm{L}=3.36 \times 10^5 \mathrm{~J}^{-1}$ 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 T K. Then, Heat gained by the ice $=$ Heat lost by the iron cube $\Rightarrow \mathrm{m} \times \mathrm{S} \times(\mathrm{T}-273)=\mathrm{M} \times \mathrm{L} \Rightarrow 8000 \mathrm{~V} \times 470 \times(\mathrm{T}-273)$ $=900 \mathrm{~V} \times\left(3.36 \times 10^5\right) \Rightarrow 376 \times 10^4 \times(\mathrm{T}-273)=3024 \times 10^5 \Rightarrow \mathrm{~T}=\frac{30240+102648}{376}$
$\Rightarrow\text{T}=\frac{132888}{376}\text{K}$
$\Rightarrow\text{T}=353.425\text{K}\approx353\text{K}$
$\Rightarrow\text{T}=353\text{K}-273\text{K}$
$\Rightarrow\text{T}=80^\circ\text{C}$
View full question & answer→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?
AnswerHeat 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.
View full question & answer→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?
AnswerGiven, 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}$
View full question & answer→Question 161 Mark
The calorie ia defined as 1cal = 4.186 joule. Why not as 1cal = 4J to make the conversions easy?
AnswerAs cal is very small unit energy may be calculated in Kcal so in this case difference between 4.186 and 4 becomes large.
View full question & answer→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}$.
AnswerHeight of the floor from which ball is dropped, $h_1 = 2.0m$ Height to which the ball rises after collision, $h_2 = 1.5m$ 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}$
View full question & answer→Question 181 Mark
A copper cube of mass $200\ g$ slides down on a rough inclined plane of inclination $37^{\circ}$ 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 $=420 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$.
AnswerMass 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$
$\Rightarrow\triangle\text{T}=0.0086=8.6\times10^{-3}{^\circ}\text{C}$
View full question & answer→Question 191 Mark
Is heat a conserved quantity?
AnswerHeat is not conserved quantity as heat only exists when there is some energy transferred.
View full question & answer→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.
AnswerGiven, 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 $kx + V\rho g = mg 200x + (2 \times 10^{-4}) \times (1000) \times (10) = 12$
$\Rightarrow\text{x}=\frac{(12-2)}{200}$
$\Rightarrow\text{x}=\frac{10}{200}=0.05\text{m}$
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$\frac{1}{2}\text{kx}^2+\text{mgh}-\text{V}\rho\text{gh}=\text{m}_1\text{s}_1\triangle\text{T}+\text{m}_2\text{s}_2\triangle\text{T}$
$\Rightarrow\frac{1}{2}\times200\times0.0025+1.2\times10\times\Big(\frac{40}{100}\Big)\\-2\times10^{-4}\times1000\times10\times\Big(\frac{40}{100}\Big)$
$\Rightarrow\Big(\frac{260}{1000}\Big)\times4200\times\triangle\text{T}+1.2\times250\times\triangle\text{T}$
$\Rightarrow0.25+4.8-0.8=1092\triangle\text{T}+300\triangle\text{T}$
$\Rightarrow1392\triangle\text{T}=4.25$
$\Rightarrow\triangle\text{T}=\frac{4.25}{1392}=0.0030531$
$\Rightarrow\triangle\text{T}=3\times10^{-3}{^\circ}\text{C}$
View full question & answer→