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(C) The heat required $Q$ is given by the integral $Q = \int m c \, dt$. Given $m = 2 \, g$,$c = 0.20 + 0.14t + 0.023t^2$,and temperature limits $t_1

Vedclass · Accepted Answer

$82$

Vedclass · Answer

(C) The heat required $Q$ is given by the integral $Q = \int m c \, dt$. Given $m = 2 \, g$,$c = 0.20 + 0.14t + 0.023t^2$,and temperature limits $t_1 = 5 \, ^\circ C$ to $t_2 = 15 \, ^\circ C$. $Q = \int_{5}^{15} 2 	imes (0.20 + 0.14t + 0.023t^2) \, dt$ $Q = 2 	imes [0.20t + 0.07t^2 + \frac{0.023}{3}t^3]_{5}^{15}$ $Q = 2 	imes [(0.20(15) + 0.07(15^2) + \frac{0.023}{3}(15^3)) - (0.20(5) + 0.07(5^2) + \frac{0.023}{3}(5^3))]$ $Q = 2 	imes [(3 + 15.75 + 25.875) - (1 + 1.75 + 0.9583)]$ $Q = 2 	imes [44.625 - 3.7083] = 2 	imes 40.9167 \approx 81.83 \, cal$. Rounding to the nearest integer,we get $82 \, cal$.

The specific heat of a substance varies with temperature as $c = 0.20 + 0.14t + 0.023t^2 \, (cal/g \cdot ^\circ C)$. The amount of heat required to raise the temperature of $2 \, g$ of the substance from $5 \, ^\circ C$ to $15 \, ^\circ C$ is: (in $, cal$)

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