A calorimeter of mass 0.2 kg and specific he | vedclass

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(D) The rate of cooling is defined as the rate of loss of heat,given by $\frac{dQ}{dt} = (m_l c_l + m_c c_c) \frac{d	heta}{dt}$.Here,$m_l = 0.5 \ kg$

Vedclass · Accepted Answer

$115$

Vedclass · Answer

(D) The rate of cooling is defined as the rate of loss of heat,given by $\frac{dQ}{dt} = (m_l c_l + m_c c_c) \frac{d	heta}{dt}$.Here,$m_l = 0.5 \ kg$,$c_l = 2400 \ J/kg \cdot K$,$m_c = 0.2 \ kg$,$c_c = 900 \ J/kg \cdot K$.The change in temperature $\Delta 	heta = 60^{\circ}C - 55^{\circ}C = 5^{\circ}C$ and the time interval $\Delta t = 1 \ minute = 60 \ seconds$.Substituting the values:$\frac{dQ}{dt} = (0.5 	imes 2400 + 0.2 	imes 900) 	imes \frac{5}{60}$$\frac{dQ}{dt} = (1200 + 180) 	imes \frac{5}{60}$$\frac{dQ}{dt} = 1380 	imes \frac{1}{12} = 115 \ J/s$.

$A$ calorimeter of mass $0.2 \ kg$ and specific heat $900 \ J/kg \cdot K$ contains $0.5 \ kg$ of a liquid of specific heat $2400 \ J/kg \cdot K$. Its temperature falls from $60^{\circ}C$ to $55^{\circ}C$ in one minute. The rate of cooling is ....... $J/s$.

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