Question

$1$ kg of water is heated from $40 ^\circ C$ to $70 ^\circ C$, If its volume remains constant, then the change in internal energy is (specific heat of water $= 4148 \,J \,kg^{-1}K^{-1})$

• A. $2.44\times 10^{5}\,J$
• B. $1.62\times 10^{5}\,J$
• C. $1.24\times 10^{5}\,J$
• D. $2.62\times 10^{5}\,J$

Answer 1

Answer: (C)

Since volume of water remains constant, then work done
$\Delta W = P dV = 0$
According to first pair of thermodynamics
$dQ = dU + dW, dU - dQ = ms \Delta T $
$= 1 \times 4148 \times (70 - 40) = 4148 \times 30 $
$= 124440\,J$
$ = 1.244 \times 10^5\,J$