Two liquids A and B are at 32,^{circ}C and 24 | vedclass

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(C) According to the principle of calorimetry,heat lost by the hotter body equals heat gained by the colder body.Let $m$ be the mass of each liquid,$c

Vedclass · Accepted Answer

$1:1$

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(C) According to the principle of calorimetry,heat lost by the hotter body equals heat gained by the colder body.Let $m$ be the mass of each liquid,$c_A$ and $c_B$ be their specific heats,and $T$ be the final mixture temperature.Heat lost by $A = m 	imes c_A 	imes (T_A - T)$Heat gained by $B = m 	imes c_B 	imes (T - T_B)$Given $T_A = 32\,^{\circ}C$,$T_B = 24\,^{\circ}C$,and $T = 28\,^{\circ}C$.Since $m_A = m_B = m$,we have:$m 	imes c_A 	imes (32 - 28) = m 	imes c_B 	imes (28 - 24)$$c_A 	imes 4 = c_B 	imes 4$$\frac{c_A}{c_B} = \frac{4}{4} = 1:1$

Two liquids $A$ and $B$ are at $32\,^{\circ}C$ and $24\,^{\circ}C$. When mixed in equal masses,the temperature of the mixture is found to be $28\,^{\circ}C$. Their specific heats are in the ratio of:

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