50 text{ g} of copper is heated to increase i | vedclass

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(C) Given:For copper: Mass $m_1 = 50 	ext{ g} = 0.05 	ext{ kg}$,Temperature rise $\Delta t_1 = 10^{\circ} 	ext{C}$,Specific heat $s_1 = 420 	ext{

Vedclass · Accepted Answer

$5$

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(C) Given:For copper: Mass $m_1 = 50 	ext{ g} = 0.05 	ext{ kg}$,Temperature rise $\Delta t_1 = 10^{\circ} 	ext{C}$,Specific heat $s_1 = 420 	ext{ J kg}^{-1} {}^{\circ} 	ext{C}^{-1}$.For water: Mass $m_2 = 10 	ext{ g} = 0.01 	ext{ kg}$,Specific heat $s_2 = 4200 	ext{ J kg}^{-1} {}^{\circ} 	ext{C}^{-1}$,Let the rise in temperature be $\Delta t_2$.Since the same amount of heat $Q$ is supplied to both,we have $Q_1 = Q_2$.Using the formula $Q = m s \Delta t$,we get:$m_1 s_1 \Delta t_1 = m_2 s_2 \Delta t_2$Substituting the values:$0.05 	imes 420 	imes 10 = 0.01 	imes 4200 	imes \Delta t_2$$210 = 42 	imes \Delta t_2$$\Delta t_2 = \frac{210}{42} = 5^{\circ} 	ext{C}$.Thus,the rise in temperature of water is $5^{\circ} 	ext{C}$.

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