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(C) According to Newton's law of cooling,the rate of heat loss is given by $\frac{dQ}{dt} = mc \frac{d	heta}{dt}$.Since the conditions are identical,

Vedclass · Accepted Answer

$8/15$

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(C) According to Newton's law of cooling,the rate of heat loss is given by $\frac{dQ}{dt} = mc \frac{d	heta}{dt}$.Since the conditions are identical,the rate of heat loss $\frac{dQ}{dt}$ is the same for both liquids.Therefore,$m_1 s_1 \frac{d	heta}{t_1} = m_2 s_2 \frac{d	heta}{t_2}$.Since the volumes are equal $(V_1 = V_2 = V)$,we can write $m = ho V$.Substituting this,we get $ho_1 V s_1 / t_1 = ho_2 V s_2 / t_2$.This simplifies to $\frac{ho_1}{ho_2} = \frac{s_2}{s_1} 	imes \frac{t_1}{t_2}$.Given $s_1 : s_2 = 3 : 4$,so $s_2 / s_1 = 4/3$.Given $t_1 = 324 \; sec$ and $t_2 = 810 \; sec$.Ratio $\frac{ho_1}{ho_2} = \frac{4}{3} 	imes \frac{324}{810} = \frac{4}{3} 	imes \frac{2}{5} = \frac{8}{15}$.

Two liquids of equal volume are cooled under identical conditions from $60^{\circ}C$ to $50^{\circ}C$ in $324 \; sec$ and $810 \; sec$ respectively. If the ratio of their specific heats is $3:4$,find the ratio of their densities. (The water equivalent of the calorimeter is negligible.)

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