A calorimeter contains 10 g of water at 20^{ | vedclass

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(C) According to Newton's Law of Cooling,the rate of heat loss is proportional to the temperature difference between the body and the surroundings: $\

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$10$

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(C) According to Newton's Law of Cooling,the rate of heat loss is proportional to the temperature difference between the body and the surroundings: $\frac{dQ}{dt} = k(T - T_s)$.Here,the heat lost by the system (calorimeter + water) is $\Delta Q = (m_w + w)c \Delta T$,where $w$ is the water equivalent of the calorimeter.The rate of cooling is $\frac{\Delta Q}{\Delta t} = \frac{(m_w + w)c \Delta T}{\Delta t} = k(T_{avg} - T_s)$.Since the temperature change $\Delta T = 5^{\circ}C$ and the average temperature $T_{avg} = 17.5^{\circ}C$ are the same in both cases,the rate of heat loss is constant.Thus,$\frac{(m_1 + w)c \Delta T}{t_1} = \frac{(m_2 + w)c \Delta T}{t_2}$.This simplifies to $\frac{m_1 + w}{t_1} = \frac{m_2 + w}{t_2}$.Given $m_1 = 10 \ g, t_1 = 10 \ min$ and $m_2 = 20 \ g, t_2 = 15 \ min$.Substituting the values: $\frac{10 + w}{10} = \frac{20 + w}{15}$.$15(10 + w) = 10(20 + w) \implies 150 + 15w = 200 + 10w$.$5w = 50 \implies w = 10 \ g$.

$A$ calorimeter contains $10 \ g$ of water at $20^{\circ}C$. The temperature falls to $15^{\circ}C$ in $10 \ min$. When the calorimeter contains $20 \ g$ of water at $20^{\circ}C$,it takes $15 \ min$ for the temperature to become $15^{\circ}C$. The water equivalent of the calorimeter is: (in $g$)

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