100 g of water is heated from 30^circ C to 5 | vedclass

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(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.Since the expansion of water is ignored,the change in volume $\Delta V = 0$

Vedclass · Accepted Answer

$8.4$

Vedclass · Answer

(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.Since the expansion of water is ignored,the change in volume $\Delta V = 0$,which implies that the work done $W = P \Delta V = 0$.Therefore,the change in internal energy $\Delta U$ is equal to the heat supplied $\Delta Q$.The formula for heat supplied is $\Delta Q = m c \Delta T$.Given: mass $m = 100 \ g = 0.1 \ kg$,specific heat $c = 4184 \ J/kg/K$,and change in temperature $\Delta T = 50^\circ C - 30^\circ C = 20 \ K$.Substituting the values: $\Delta U = 0.1 \ kg 	imes 4184 \ J/kg/K 	imes 20 \ K = 8368 \ J$.Converting to $kJ$: $\Delta U = 8368 / 1000 \approx 8.4 \ kJ$.

$100 \ g$ 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 ...... $kJ$ (specific heat of water is $4184 \ J/kg/K$).

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