Heat and Thermodynamics

JEE Physics · 315 questions · Page 32 of 32 · Click an option or "Show Solution" to reveal answer

Q311

A cylinder with fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20°C is : [Given that R = 8.31 J mol–1 K–1]

A 374 J

B 700 J

C 748 J

D 350 J

Correct Answer

Option C

Solution

\Delta Q = n{C_v}

\Delta T = n{3 \over 2}R\Delta T

= \left( {{{67.2} \over {22.4}}} \right)\left( {{3 \over 2} \times 8.31} \right)\left( {20} \right)

\approx 748\,J

Q312

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by

\Delta

T. The amount of heat absorbed by gas is (R is gas constant) :

{1 \over 2}

\Delta

{1 \over 2}

\Delta

{3 \over 2}

\Delta

{2K \over 3}

\Delta

Correct Answer

Option B

Solution

VT = K $\Rightarrow$ V

\left( {{{PV} \over {nR}}} \right)

= k $\Rightarrow$ PV2 = K $\because$ C =

{R \over {1 - x}} +

Cv (For polytropic process) C =

{R \over {1 - 2}} + {{3R} \over 2}

{R \over 2}

$\therefore$

\Delta

Q = nC

\Delta

T =

{R \over 2} \times \Delta

Q313

Ice at –20oC is added to 50 g of water at 40oC. When the temperature of the mixture reaches 0oC, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2J/g/oC Specific heat of Ice = 2.1J/g/oC Heat of fusion of water at 0oC= 334J/g)

A 100 g

B 60 g

C 50 g

D 40 g

Correct Answer

Option D

Solution

Let amount of ice is m gm.

According to principal of calorimeter heat taken by ice = heat given by water $\therefore$ 20 $\times$ 2.1 $\times$ m + (m $-$ 20) $\times$ 334 = 50 $\times$ 4.2 $\times$ 40 376 m = 8400 + 6680 m = 40.1

Q314

One

kg

of a diatomic gas is at a pressure of

8 \times {10^4}\,N/{m^2}.

The density of the gas is

4kg/{m^3}

. What is the energy of the gas due to its thermal motion ?

5 \times {10^4}\,J

6 \times {10^4}\,J

7 \times {10^4}\,J

3 \times {10^4}\,J

Correct Answer

Option A

Solution

Volume\,\, = \,\,{{mass} \over {density}} = {1 \over 4}{m^3}

K.E = {5 \over 2}PV

= {5 \over 2} \times 8 \times {10^4} \times {1 \over 4}

= 5 \times {10^4}J

Q315

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75oC. T is given by: (Given : room temperature = 30oC, specific heat of copper = 0.1 cal/gmoC)

A 825oC

B 800oC

C 885oC

D 1250oC

Correct Answer

Option C

Solution

According to principle of calorimetry, Heat lost = Heat gain 100 × 0.1( – 75) = 100 × 0.1 × 45 + 170 × 1 × 45 10 – 750 = 450 + 7650 10 = 1200 + 7650 = 8850 T = 885°C

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