A metal block is made from a mixture of 2.4 | vedclass

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(D) The total heat supplied $Q$ is given by the formula: $Q = (m_1 s_1 + m_2 s_2 + m_3 s_3) \Delta T$. Here,$m_1 = 2.4 \ kg$,$s_1 = 0.216 \ cal \cdot

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$80$

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(D) The total heat supplied $Q$ is given by the formula: $Q = (m_1 s_1 + m_2 s_2 + m_3 s_3) \Delta T$. Here,$m_1 = 2.4 \ kg$,$s_1 = 0.216 \ cal \cdot kg^{-1} \cdot ^{\circ}C^{-1}$ (Aluminium). $m_2 = 1.6 \ kg$,$s_2 = 0.0917 \ cal \cdot kg^{-1} \cdot ^{\circ}C^{-1}$ (Brass). $m_3 = 0.8 \ kg$,$s_3 = 0.0931 \ cal \cdot kg^{-1} \cdot ^{\circ}C^{-1}$ (Copper). $Q = 44.4 \ cal$. Substituting the values: $44.4 = (2.4 	imes 0.216 + 1.6 	imes 0.0917 + 0.8 	imes 0.0931) \Delta T$. $44.4 = (0.5184 + 0.14672 + 0.07448) \Delta T$. $44.4 = (0.7396) \Delta T$. $\Delta T = 44.4 / 0.7396 \approx 60^{\circ} C$. Since $\Delta T = T_{final} - T_{initial}$,we have $60 = T_{final} - 20$. Therefore,$T_{final} = 80^{\circ} C$.

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