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

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(A) The amount of heat supplied is the same for both copper and water. According to the principle of calorimetry,$Q = mc\Delta T$.Since the heat suppl

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$5$

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(A) The amount of heat supplied is the same for both copper and water. According to the principle of calorimetry,$Q = mc\Delta T$.Since the heat supplied is equal,we have: $m_c c_c \Delta T_c = m_w c_w \Delta T_w$.Given:$m_c = 50 \, g = 50 	imes 10^{-3} \, kg$$c_c = 420 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$$\Delta T_c = 10 \, ^\circ C$$m_w = 10 \, g = 10 	imes 10^{-3} \, kg$$c_w = 4200 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$Substituting the values:$(50 	imes 10^{-3}) 	imes 420 	imes 10 = (10 	imes 10^{-3}) 	imes 4200 	imes \Delta T_w$$50 	imes 420 	imes 10 = 10 	imes 4200 	imes \Delta T_w$$210000 = 42000 	imes \Delta T_w$$\Delta T_w = \frac{210000}{42000} = 5 \, ^\circ C$.Thus,the rise in temperature of water is $5 \, ^\circ C$.

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