A sphere of 0.047 ; kg aluminium is placed fo | vedclass

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(C) According to the principle of calorimetry,at steady state: Heat lost by aluminium sphere = Heat gained by water + Heat gained by calorimeter.Mass

Vedclass · Accepted Answer

$0.91$

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(C) According to the principle of calorimetry,at steady state: Heat lost by aluminium sphere = Heat gained by water + Heat gained by calorimeter.Mass of aluminium sphere $(m_1) = 0.047 \; kg$.Initial temperature of aluminium sphere = $100 \; ^{\circ}C$.Final temperature = $23 \; ^{\circ}C$.Change in temperature $(\Delta T_1) = 100 \; ^{\circ}C - 23 \; ^{\circ}C = 77 \; ^{\circ}C$.Heat lost by aluminium sphere $Q_{lost} = m_1 s_{Al} \Delta T_1 = 0.047 	imes s_{Al} 	imes 77$.Mass of water $(m_2) = 0.25 \; kg$.Mass of copper calorimeter $(m_3) = 0.14 \; kg$.Initial temperature of water and calorimeter = $20 \; ^{\circ}C$.Final temperature = $23 \; ^{\circ}C$.Change in temperature $(\Delta T_2) = 23 \; ^{\circ}C - 20 \; ^{\circ}C = 3 \; ^{\circ}C$.Specific heat of water $(s_w) = 4.18 	imes 10^3 \; J \; kg^{-1} K^{-1}$.Specific heat of copper $(s_{cu}) = 0.386 	imes 10^3 \; J \; kg^{-1} K^{-1}$.Heat gained by water and calorimeter $Q_{gained} = (m_2 s_w + m_3 s_{cu}) \Delta T_2$.$Q_{gained} = (0.25 	imes 4.18 	imes 10^3 + 0.14 	imes 0.386 	imes 10^3) 	imes 3$.$Q_{gained} = (1045 + 54.04) 	imes 3 = 1099.04 	imes 3 = 3297.12 \; J$.Equating heat lost and gained:$0.047 	imes s_{Al} 	imes 77 = 3297.12$.$3.619 	imes s_{Al} = 3297.12$.$s_{Al} = 3297.12 / 3.619 \approx 911.06 \; J \; kg^{-1} K^{-1}$.Converting to $kJ \; kg^{-1} K^{-1}$,$s_{Al} \approx 0.911 \; kJ \; kg^{-1} K^{-1}$.

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