A mass of 50,g of water in a closed vessel,wi | vedclass

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(D) According to Newton's law of cooling,the rate of heat loss $\frac{dQ}{dt}$ depends on the temperature difference between the body and the surround

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$0.5$

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(D) According to Newton's law of cooling,the rate of heat loss $\frac{dQ}{dt}$ depends on the temperature difference between the body and the surroundings. Since the vessel,surroundings,and temperature range are identical,the rate of heat loss is the same for both cases.Let $m_w = 50\,g$ be the mass of water,$C_w = 1\,cal/(g \cdot ^oC)$ be the specific heat of water,$m_l = 100\,g$ be the mass of the liquid,$C_l$ be the specific heat of the liquid,and $W = 30\,g$ be the water equivalent of the vessel.The total heat lost is given by $Q = (mC + W) \Delta T$.Since the time $t$ is the same,the rate of heat loss is:$\frac{(m_w C_w + W) \Delta T}{t} = \frac{(m_l C_l + W) \Delta T}{t}$Canceling $\frac{\Delta T}{t}$ from both sides:$m_w C_w + W = m_l C_l + W$$m_w C_w = m_l C_l$Substituting the values:$50 	imes 1 = 100 	imes C_l$$C_l = \frac{50}{100} = 0.5\,cal/(g \cdot ^oC) = 0.5\,kcal/(kg \cdot ^oC)$.

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