2,g of steam condenses when passed through 40 | vedclass

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Question

(A) According to the principle of calorimetry,Heat gained by water = Heat lost by steam.Heat gained by water $(Q_{gain})$ = $m_w \cdot s_w \cdot \Delt

Vedclass · Accepted Answer

$540$

Vedclass · Answer

(A) According to the principle of calorimetry,Heat gained by water = Heat lost by steam.Heat gained by water $(Q_{gain})$ = $m_w \cdot s_w \cdot \Delta T_w$$Q_{gain} = 40\,g 	imes 1\,cal/g^{\circ}C 	imes (54.3^{\circ}C - 25^{\circ}C) = 40 	imes 29.3 = 1172\,cal$.Heat lost by steam $(Q_{lost})$ = Heat released during condensation + Heat released by condensed water cooling down.$Q_{lost} = m_s \cdot L + m_s \cdot s_w \cdot \Delta T_s$$Q_{lost} = 2\,g 	imes L + 2\,g 	imes 1\,cal/g^{\circ}C 	imes (100^{\circ}C - 54.3^{\circ}C)$$Q_{lost} = 2L + 2 	imes 45.7 = 2L + 91.4$.Equating the two:$1172 = 2L + 91.4$$2L = 1172 - 91.4 = 1080.6$$L = 540.3\,cal/g$.Rounding to the nearest option,the latent heat is approximately $540\,cal/g$.

$2\,g$ of steam condenses when passed through $40\,g$ of water initially at $25\,^{\circ}C$. The condensation of steam raises the temperature of water to $54.3\,^{\circ}C$. What is the latent heat of steam in $cal/g$?

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