Question

in a processat constant volume 24cal/gram of heat has been added to a system .calculate the change in its internal energy​

Answer 1

Explanation:

heat = ΔQ=Mass × specific heat×temperature change=15×0.2×5 cal=15 cal

Also, isochoric process implies work done =0

Applying first law, Change in internal energy= ΔQ =15 cal

Answer 2

Concept:

• Thermodynamics
• Isochoric process
• An isochoric process in thermodynamics is a thermodynamic process in which the volume of the closed system experiencing the process remains constant.
• It is also known as a constant-volume process, isovolumetric process, or isometric process.
• Internal energy
• Work done
• The change in internal energy is the sum of the heat and the work done

Given:

• Process in which volume is constant
• isochoric process
• Heat added to the system Q = 24 cal/g

Find:

• The change in internal energy ΔU

Solution:

According to the first law of thermodynamics, we know that ΔU = Q+W

W = 0, as the process is isochoric and there is no change in volume

ΔU = Q+ 0

ΔU = Q

Q = 24

ΔU = Q = 24

ΔU = 24

The change in internal energy is 24 J.

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