$100\ g$ of water is heated from $30^o\ C$ to $50^o\ C$ Ignoring the slight expansion of the water, the change in its internal energy is ...... $kJ$ (specific heat of water is $4184\ J/kg/K$)
Medium
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As work done $=0$
$[\mathrm{as} W=p \Delta V \text { and } \Delta V=0, \text { so } W=0]$
$\Delta U=m C \Delta T$
$=100 \times 10^{-3} \times 4184 \times(50-30) \approx 8.4 \mathrm{kJ}$
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