A calorimeter of water equivalent 20 , g cont | vedclass

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Question

(B) The principle of calorimetry states that heat lost by the hot body equals heat gained by the cold body.Heat gained by the calorimeter and water:$Q

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The principle of calorimetry states that heat lost by the hot body equals heat gained by the cold body.Heat gained by the calorimeter and water:$Q_{	ext{gained}} = (m_{	ext{cal}} + m_{	ext{water}}) 	imes c_w 	imes \Delta T$$Q_{	ext{gained}} = (20 \, g + 180 \, g) 	imes 1 \, 	ext{cal} \, g^{-1} {}^{\circ} C^{-1} 	imes (31^{\circ} C - 25^{\circ} C)$$Q_{	ext{gained}} = 200 	imes 6 = 1200 \, 	ext{cal}$.Heat lost by steam at $100^{\circ} C$ to become water at $31^{\circ} C$:$Q_{	ext{lost}} = m 	imes L_v + m 	imes c_w 	imes \Delta T$$Q_{	ext{lost}} = m 	imes 540 + m 	imes 1 	imes (100^{\circ} C - 31^{\circ} C)$$Q_{	ext{lost}} = m 	imes (540 + 69) = 609m$.Equating heat gained and lost:$1200 = 609m$$m = \frac{1200}{609} \approx 1.97 \, g$.Thus,the value of '$m$' is close to $2 \, g$.

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