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(A) The heat energy supplied to the water is given by the formula $Q = m c \Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\

Vedclass · Accepted Answer

$1.25 	imes 10^5 \ J$

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(A) The heat energy supplied to the water is given by the formula $Q = m c \Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.Given: $m = 500 \ g = 0.5 \ kg$, $c = 4184 \ J \ kg^{-1} \ K^{-1}$, $\Delta T = 90^{\circ}C - 30^{\circ}C = 60 \ K$.Substituting the values: $Q = 0.5 	imes 4184 	imes 60$.$Q = 0.5 	imes 251040 = 125520 \ J$.Since the volume change for liquid water during heating is negligible, the work done is approximately zero, and the change in internal energy $\Delta U \approx Q$.Therefore, $\Delta U = 1.2552 	imes 10^5 \ J \approx 1.25 	imes 10^5 \ J$.

$500 \ g$ of water is heated from $30^{\circ}C$ to $90^{\circ}C$. What is the change in its internal energy? (Specific heat of water is $4184 \ J \ kg^{-1} \ K^{-1}$)

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