The specific heat capacity of a substance is | vedclass

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(D) The initial temperature is $T_i = -73^{\circ} C = (-73 + 273) \ K = 200 \ K$.The final temperature is $T_f = 27^{\circ} C = (27 + 273) \ K = 300 \

Vedclass · Accepted Answer

$25000$

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(D) The initial temperature is $T_i = -73^{\circ} C = (-73 + 273) \ K = 200 \ K$.The final temperature is $T_f = 27^{\circ} C = (27 + 273) \ K = 300 \ K$.The heat required $Q$ is given by the integral $Q = \int_{T_i}^{T_f} m C dT$.Given $m = 1 \ kg$ and $C = kT$,we have $Q = \int_{200}^{300} (1) (kT) dT$.$Q = k \int_{200}^{300} T dT = k \left[ \frac{T^2}{2} \right]_{200}^{300}$.$Q = \frac{k}{2} [300^2 - 200^2] = \frac{k}{2} [90000 - 40000] = \frac{k}{2} [50000]$.$Q = 25000 k$.Comparing this with $nk$,we get $n = 25000$.

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