Question

$100\text{g}$ of water is heated from ${30}^{\circ }C$ to ${50}^{\circ }C$. Ignoring the slight expansion of the water, the change in its internal energy is $\left(\text{in kJ}\right)$ (specific heat of water is $4184\text{J/kg/K})$

Answer 1

Given mass of water $m=100g=0.1kg$
Specific heat of water $S=4184{\text{J kg}}^{-1}{K}^{-1}$
Change in temperature $\Delta T={50}^{\circ }C–{30}^{\circ }C={20}^{\circ }C$

From first law of thermodynamics
$\Delta Q=\Delta U+\Delta W$
Ignoring expansion $\Delta W=0$
$\Delta U=ms\Delta T=0.1\times 4.184\times 20=8.368\text{kJ}$
$\Delta U\approx 8.4\text{kJ}$