In a calorimeter of water equivalent 20 ,g,wa | vedclass

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Question

(B) Let the mass of steam that condenses be $x \,g$.The initial temperature of water is $288 \,K = 15^{\circ} C$.The final temperature of water is $15

Vedclass · Accepted Answer

$11.7$

Vedclass · Answer

(B) Let the mass of steam that condenses be $x \,g$.The initial temperature of water is $288 \,K = 15^{\circ} C$.The final temperature of water is $15^{\circ} C + 6.5^{\circ} C = 21.5^{\circ} C$.Heat released by steam in condensing at $100^{\circ} C$ and cooling down to $21.5^{\circ} C$:$Q_{	ext{released}} = m_s L_v + m_s c_w \Delta T_s$$Q_{	ext{released}} = x 	imes 540 + x 	imes 1 	imes (100 - 21.5) = 540x + 78.5x = 618.5x \,cal$.Heat absorbed by water and calorimeter:$Q_{	ext{absorbed}} = (m_w c_w + W) \Delta T_w$$Q_{	ext{absorbed}} = (1100 	imes 1 + 20) 	imes 6.5 = 1120 	imes 6.5 = 7280 \,cal$.By the principle of calorimetry,$Q_{	ext{released}} = Q_{	ext{absorbed}}$:$618.5x = 7280$$x = \frac{7280}{618.5} \approx 11.77 \,g$.Rounding to the nearest provided option,$x = 11.7 \,g$.

In a calorimeter of water equivalent $20 \,g$,water of mass $1.1 \,kg$ is taken at $288 \,K$ temperature. If steam at temperature $373 \,K$ is passed through it and the temperature of the water increases by $6.5^{\circ} C$,then the mass of steam condensed is ............ $g$.

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