In Newton's experiment of cooling,the water e | vedclass

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(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T - T_0)$.For a calorimeter,the rate of heat loss is $\fra

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T - T_0)$.For a calorimeter,the rate of heat loss is $\frac{dQ}{dt} = (mC + W) \frac{dT}{dt}$,where $m$ is the mass,$C$ is the specific heat,and $W$ is the water equivalent.Since the cooling range $(70^{\circ}C$ to $60^{\circ}C)$ and the surrounding temperature are the same,the rate of heat loss $\frac{dQ}{dt}$ is constant for both.Thus,$\frac{(m_W C_W + W) \Delta T}{t_W} = \frac{(m_L C_L + W) \Delta T}{t_L}$.Rearranging for $C_L$: $C_L = \frac{1}{m_L} \left[ \frac{t_L}{t_W} (m_W C_W + W) - W ight]$.Given: $m_W = 350 \ g$,$C_W = 1 \ Cal/g^{\circ}C$,$W = 10 \ g$,$t_W = 3 \ min = 180 \ s$,$m_L = 300 \ g$,$t_L = 95 \ s$.Substituting the values: $C_L = \frac{1}{300} \left[ \frac{95}{180} (350 	imes 1 + 10) - 10 ight]$.$C_L = \frac{1}{300} \left[ \frac{95}{180} (360) - 10 ight] = \frac{1}{300} [95 	imes 2 - 10] = \frac{190 - 10}{300} = \frac{180}{300} = 0.6 \ Cal/g^{\circ}C$.

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