The internal energy change in a system that has absorbed 2 kcal of heat and done 500 ~J of work is

Correct answer: D. 7900 ~J Key Idea Heat given to a system ( Δ Q ) is equal to the sum of increase in the internal energy (Δ u) and the work done (Δ W) by the system against the surrounding and 1 cal=4.2 J . According to first law of thermodynamics aligned Δ U& =Q-W & =2 ×4.2 ×1000-500 & =8400-500 & =7900 J aligned

NEET 2009

Exam: NEET
Year: 2009
Subject: Physics
Chapter: Thermodynamics

NEET 2009 · Physics · Thermodynamics

The internal energy change in a system that has absorbed 2 kcal of...

The internal energy change in a system that has absorbed

2 \mathrm{kcal}

of heat and done

500 \mathrm{~J}

of work is

Options

A $8900 \mathrm{~J}$
B $6400 \mathrm{~J}$
C $5400 \mathrm{~J}$
D $7900 \mathrm{~J}$

Correct answer

D. $7900 \mathrm{~J}$

Step-by-step solution

Key Idea Heat given to a system (

\Delta Q

) is equal to the sum of increase in the internal energy

(\Delta \mathrm{u})

and the work done

(\Delta W)

by the system against the surrounding and

1 \mathrm{cal}=4.2 \mathrm{ J}

. According to first law of thermodynamics

\begin{aligned} \Delta \mathrm{U}& =\mathrm{Q}-\mathrm{W}\\ & =2 \times4.2 \times1000-500 \\ & =8400-500 \\ & =7900 \mathrm{ J} \end{aligned}

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