A 10.0 ,W electrical heater is used to heat a | vedclass

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(A) Let $C$ be the heat capacity of the container $(C = m_c s_c)$.Case $1$: Water in the container.Energy supplied = $P 	imes t_1 = 10 	imes (15 	i

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) Let $C$ be the heat capacity of the container $(C = m_c s_c)$.Case $1$: Water in the container.Energy supplied = $P 	imes t_1 = 10 	imes (15 	imes 60) = 9000 \,J$.Heat absorbed = $(m_w s_w + C) \Delta T_1 = (0.5 	imes 4200 + C) 	imes 3 = 6300 + 3C$.Equating energy: $6300 + 3C = 9000 \Rightarrow 3C = 2700 \Rightarrow C = 900 \,J K^{-1}$.Case $2$: Oil in the container.Energy supplied = $P 	imes t_2 = 10 	imes (20 	imes 60) = 12000 \,J$.Heat absorbed = $(m_o s_o + C) \Delta T_2 = (2 	imes s_o + 900) 	imes 2 = 4s_o + 1800$.Equating energy: $4s_o + 1800 = 12000 \Rightarrow 4s_o = 10200 \Rightarrow s_o = 2550 \,J K^{-1} kg^{-1}$.Expressing in terms of $10^3 \,J K^{-1} kg^{-1}$,we get $s_o = 2.55 	imes 10^3 \,J K^{-1} kg^{-1}$. Given the options,the closest value is $2.5$.

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