2,kg of metal at 100,^{circ}C is cooled by 1, | vedclass

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Question

(A) According to the principle of calorimetry,heat lost by the metal equals heat gained by the water.Let $m_1 = 2\,kg$ be the mass of the metal and $s

Vedclass · Accepted Answer

$50\,^{\circ}C$

Vedclass · Answer

(A) According to the principle of calorimetry,heat lost by the metal equals heat gained by the water.Let $m_1 = 2\,kg$ be the mass of the metal and $s_1$ be its specific heat capacity.Let $m_2 = 1\,kg$ be the mass of the water and $s_2$ be its specific heat capacity.Given $s_1 = \frac{1}{2} s_2$.Let $	heta$ be the final equilibrium temperature.Heat lost by metal $= m_1 s_1 (100 - 	heta) = 2 	imes (\frac{s_2}{2}) 	imes (100 - 	heta) = s_2 (100 - 	heta)$.Heat gained by water $= m_2 s_2 (	heta - 0) = 1 	imes s_2 	imes 	heta = s_2 	heta$.Equating the two: $s_2 (100 - 	heta) = s_2 	heta$.$100 - 	heta = 	heta$.$2	heta = 100$.$	heta = 50\,^{\circ}C$.

$2\,kg$ of metal at $100\,^{\circ}C$ is cooled by $1\,kg$ of water at $0\,^{\circ}C$. If the specific heat capacity of the metal is $\frac{1}{2}$ of the specific heat capacity of water,the final temperature of the mixture would be:

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