Q: If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is

100^\circC. Heat to melt 1 g ice at 0^\circC: Q_1=mL=1\times80=80 cal. Heat to raise 1 g water from 0^\circC to 100^\circC: Q_2=mc\Delta\theta=1\times1\times(100-0)=100 cal. Total Q=Q_1+Q_2=180 cal. Heat released by 1 g steam condensing at 100^\circC: Q'=mL'=1\times536=536 cal. Since 536>180, not all the steam condenses and the mixture settles at 100^\circC.

Q: 10 g of ice cubes at 0^\circC are released in a tumbler (water equivalent 55 g) at 40^\circC. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L=80 cal/g)

22^\circC. Let \theta be the equilibrium temperature. Heat gained by ice =mL+ms(\theta-0)=10\times80+10\times1\times\theta. Heat lost by the tumbler and its contents =55\times(40-\theta). Equating: 800+10\theta=55(40-\theta)\Rightarrow65\theta=1400\Rightarrow\theta=\dfrac{1400}{65}\approx22^\circC.

Question 1

The quantities of heat required to raise the temperature of two solid copper spheres of radii r_1 and r_2 (r_1=1.5r_2) through 1 K are in the ratio

Accepted Answer

\dfrac{27}{8}. Heat required, Q=mc\Delta T=\left(\dfrac{4}{3}\pi r^3\rho\right)c\Delta T (since m=V_{sphere}\,\rho). As \pi,\rho,c and T are constants, Q\propto r^3, so \dfrac{Q_1}{Q_2}=\left(\dfrac{r_1}{r_2}\right)^3=(1.5)^3=\dfrac{27}{8}.

Question 2

A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is [Latent heat of ice is 3.4	imes10^5 J/kg and g=10 N/kg]

Accepted Answer

136 km. By conservation of energy, energy gained during the fall from height h is E=mgh. Only one-quarter is absorbed by the ice, so \dfrac{mgh}{4}=mL_f\Rightarrow h=\dfrac{4L_f}{g}=\dfrac{4	imes3.4	imes10^5}{10}=13.6	imes10^4\,	ext{m}=136\,	ext{km}.

Question 3

Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100°C, while the other one is at 0°C. If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is

Accepted Answer

more than 50°C. Heat lost by the hotter body equals heat gained by the colder body. Because heat capacity increases with temperature, the body at 100°C has a greater heat capacity than the body at 0°C, so the equilibrium temperature is shifted towards 100°C, i.e. T_c>50°C (more than 50°C).

Question 4

Steam at 100^\circC is passed into 20 g of water at 10^\circC. When water acquires a temperature of 80^\circC, the mass of water present will be [Take specific heat of water =1 cal g^{-1} ^\circC^{-1} and latent heat of steam =540 cal g^{-1}]

Accepted Answer

22.5 g. By calorimetry, heat lost by steam = heat gained by water. Let m' be the steam that condenses: m'L+m's\Delta T=ms\Delta t\Rightarrow m'	imes540+m'	imes1	imes(100-80)=20	imes1	imes(80-10)\Rightarrow m'=\dfrac{20	imes70}{560}=2.5 g. Net mass of water =20+2.5=22.5 g.

Question 5

Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which one of the following graphs represents the variation of temperature with time?

Accepted Answer

(a). With a constant rate of heat supply, within a single phase Q=mc\Delta T so the temperature rises uniformly, while during a change of state the temperature stays constant, giving a horizontal plateau. The curve that rises, then holds flat during the phase change, then rises again (graph a) represents this behaviour.

Question 6

When 1 kg of ice at 0^\circC melts to water at 0^\circC, the resulting change in its entropy, taking latent heat of ice to be 80 cal/^\circC, is

Accepted Answer

293 cal/K. Change in entropy, \Delta S=\dfrac{ml}{T}=\dfrac{1000	imes80}{273}=293 cal K^{-1}.

Question 7

If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is

Accepted Answer

100^\circC. Heat to melt 1 g ice at 0^\circC: Q_1=mL=1	imes80=80 cal. Heat to raise 1 g water from 0^\circC to 100^\circC: Q_2=mc\Delta	heta=1	imes1	imes(100-0)=100 cal. Total Q=Q_1+Q_2=180 cal. Heat released by 1 g steam condensing at 100^\circC: Q'=mL'=1	imes536=536 cal. Since 536>180, not all the steam condenses and the mixture settles at 100^\circC.

Question 8

Thermal capacity of 40 g of aluminium (s=0.2 cal/g-K) is

Accepted Answer

33.6 J/^\circC. Thermal capacity is the heat needed to raise the temperature of the whole body by 1 K, i.e. =ms=40	imes0.2=8 cal/^\circC =4.2	imes8=33.6 J/^\circC.

Question 9

Two containers A and B are partly filled with water and closed. The volume of A is twice that of B and it contains half the amount of water in B. If both are at the same temperature, the water vapour in the containers will have pressure in the ratio of

Accepted Answer

1:1. Vapour pressure of a substance is independent of the amount of substance; it depends only on temperature. Since both containers are at the same temperature, the ratio is 1:1.

Question 10

10 g of ice cubes at 0^\circC are released in a tumbler (water equivalent 55 g) at 40^\circC. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L=80 cal/g)

Accepted Answer

22^\circC. Let 	heta be the equilibrium temperature. Heat gained by ice =mL+ms(	heta-0)=10	imes80+10	imes1	imes	heta. Heat lost by the tumbler and its contents =55	imes(40-	heta). Equating: 800+10	heta=55(40-	heta)\Rightarrow65	heta=1400\Rightarrow	heta=\dfrac{1400}{65}\approx22^\circC.

Specific Heat Capacity, Calorimetry and Change of State — NEET Physics PYQ MCQs with Solutions

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