Density of water at 4 ^circ C and 20 ^circ C | vedclass

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(D) The heat supplied to the water is given by $Q = m S \Delta T$.Given $m = 4 \ kg$, $S = 4200 \ J/kg \cdot K$, and $\Delta T = 20 - 4 = 16 \ K$.$Q =

Vedclass · Accepted Answer

$268799.2$

Vedclass · Answer

(D) The heat supplied to the water is given by $Q = m S \Delta T$.Given $m = 4 \ kg$, $S = 4200 \ J/kg \cdot K$, and $\Delta T = 20 - 4 = 16 \ K$.$Q = 4 	imes 4200 	imes 16 = 268800 \ J$.The work done by the water during expansion is $W = P \Delta V$.The change in volume is $\Delta V = m(\frac{1}{ho_f} - \frac{1}{ho_i}) = 4(\frac{1}{998} - \frac{1}{1000}) = 4(\frac{1000 - 998}{998000}) = 4(\frac{2}{998000}) = \frac{8}{998000} \ m^3$.Given $P = 10^5 \ Pa$, the work done is $W = 10^5 	imes \frac{8}{998000} = \frac{800000}{998000} \approx 0.8016 \ J$.According to the first law of thermodynamics, $\Delta U = Q - W$.$\Delta U = 268800 - 0.8016 = 268799.1984 \ J \approx 268799.2 \ J$.

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