Steam at 100,^{circ}C is passed into 1.1,kg o | vedclass

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Question

(A) The total mass of the system (water + calorimeter) is $M = 1.1\,kg + 0.02\,kg = 1.12\,kg$.The temperature rise is $\Delta T = 80\,^{\circ}C - 15\,

Vedclass · Accepted Answer

$0.13$

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(A) The total mass of the system (water + calorimeter) is $M = 1.1\,kg + 0.02\,kg = 1.12\,kg$.The temperature rise is $\Delta T = 80\,^{\circ}C - 15\,^{\circ}C = 65\,^{\circ}C$.Heat gained by the water and calorimeter is $Q_{gain} = M \cdot c_w \cdot \Delta T$,where $c_w$ is the specific heat capacity of water.Heat lost by the steam consists of two parts: latent heat of condensation at $100\,^{\circ}C$ and heat released by the condensed water cooling from $100\,^{\circ}C$ to $80\,^{\circ}C$.Let $m$ be the mass of steam condensed. The latent heat of vaporization $L_v \approx 540\,cal/g = 540\,kcal/kg$.Using the principle of calorimetry: $Q_{gain} = Q_{lost}$.$M \cdot c_w \cdot \Delta T = m \cdot L_v + m \cdot c_w \cdot (100\,^{\circ}C - 80\,^{\circ}C)$.Using units in $kcal$ and $kg$: $1.12 	imes 1 	imes 65 = m 	imes 540 + m 	imes 1 	imes 20$.$72.8 = m(560)$.$m = 72.8 / 560 = 0.13\,kg$.

Steam at $100\,^{\circ}C$ is passed into $1.1\,kg$ of water contained in a calorimeter of water equivalent $0.02\,kg$ at $15\,^{\circ}C$ until the temperature of the calorimeter rises to $80\,^{\circ}C$. The mass of steam condensed in kilograms is:

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