7. The specific heat capacity of a body with a mass of $m=3$ kg depends on the temperature as follows: $c=c_{0}(1+\alpha t)$, where $c_{0}=200$ J/kg $ \cdot{ }^{\circ} \mathrm{C}$ is the specific heat capacity at $0^{\circ} \mathrm{C}$, $\alpha=0.05^{\circ} \mathrm{C}^{-1}$ is the temperature coefficient, and $t$ is the temperature in degrees Celsius. Determine the amount of heat that must be transferred to this body to heat it from $30^{\circ} \mathrm{C}$ to $80^{\circ} \mathrm{C}$. (10 points)
Answer: 112.5 kJ

Solution. Given that the specific heat capacity depends linearly on temperature, we can calculate its average value: $c_{c p}=\frac{c_{0}\left(1+\alpha t_{\mu}\right)+c_{0}\left(1+\alpha t_{\kappa}\right)}{2}=750 \frac{\text { J }}{\text { kg } \cdot{ }^{\circ} \mathrm{C}} \quad(5$ points $) . \quad$ The required amount of heat: $Q=c_{q p} m \Delta t=750 \cdot 3 \cdot 50=112500$ J = 112.5 kJ (5 points).