The time taken for a calorimeter containing 7 | vedclass

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(B) According to Newton's law of cooling, the rate of cooling is proportional to the heat lost per unit time. Assuming the rate of heat loss $(dQ/dt)$

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) According to Newton's law of cooling, the rate of cooling is proportional to the heat lost per unit time. Assuming the rate of heat loss $(dQ/dt)$ is constant for the same temperature range:Heat lost $Q = (w + m) C_V \Delta T$where $w$ is the water equivalent of the calorimeter, $m$ is the mass of water, $C_V$ is the specific heat of water, and $\Delta T = 62^{\circ} C - 58^{\circ} C = 4^{\circ} C$.Since the rate of cooling is constant, the time taken $t \propto Q$.Case $I$: $t_1 = 9 \ 	ext{minutes}$, $m_1 = 75 \ g$$9 \propto (w + 75) C_V 	imes 4^{\circ} \quad \dots (i)$Case $II$: $t_2 = 12 \ 	ext{minutes}$, $m_2 = 105 \ g$$12 \propto (w + 105) C_V 	imes 4^{\circ} \quad \dots (ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{9}{12} = \frac{(w + 75) C_V 	imes 4^{\circ}}{(w + 105) C_V 	imes 4^{\circ}}$$\frac{3}{4} = \frac{w + 75}{w + 105}$$3(w + 105) = 4(w + 75)$$3w + 315 = 4w + 300$$w = 315 - 300 = 15 \ g$Thus, the water equivalent of the calorimeter is $15 \ g$.

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