50,g of copper is heated to increase its temp | vedclass

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(A) The heat supplied to the copper is given by $Q = m_C c_C \Delta 	heta_C$.Given: $m_C = 50\,g = 0.05\,kg$,$c_C = 420\,J\,kg^{-1}\,^{\circ}C^{-1}$,

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$5$

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(A) The heat supplied to the copper is given by $Q = m_C c_C \Delta 	heta_C$.Given: $m_C = 50\,g = 0.05\,kg$,$c_C = 420\,J\,kg^{-1}\,^{\circ}C^{-1}$,$\Delta 	heta_C = 10\,^{\circ}C$.$Q = 0.05 	imes 420 	imes 10 = 210\,J$.Now,the same quantity of heat $Q$ is given to $10\,g$ of water $(m_W = 0.01\,kg)$.Using $Q = m_W c_W \Delta 	heta_W$,where $c_W = 4200\,J\,kg^{-1}\,^{\circ}C^{-1}$:$\Delta 	heta_W = \frac{Q}{m_W c_W} = \frac{210}{0.01 	imes 4200} = \frac{210}{42} = 5\,^{\circ}C$.

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