US PRINTER VERSION BA Chapter 18, Problem 031 What mass of steam at 100°C must b

Question

US PRINTER VERSION BA Chapter 18, Problem 031 What mass of steam at 100°C must be mixed with 425 g of ice at its thermally insulated con water is 4186 3/kg-K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 ka/kg. melting point, in a thermally insulated container, to produce liquid water at 88.0°C? The specific heat df Number Units the tolerance is +/-29 Click if you would like to Show Work for this question: Open Show Wors By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor. Question Attempts: O of 1 used SAVE FOR LATER SUBNIT ANSWER Earn Maximum Points available only if you answer this question correctly in your first attempt re to search 1/21/2018

Explanation / Answer

Heat gained by ice = Heat lost by steam

Q1 = Q2

Q1 = Qa + Qb

Qa = heat required to melt the ice into water = m*Lf

Qb = heat required to increase temperature of water from 0 C to 88 C = m*C*dT1

Q2 = Qc + Qd

Qc = Heat required to change 100 C steam into 100 C water = M*Lv

Qd = heat required to decrease temperature of water from 100 C to 88 C = M*C*dT2

So,

m*Lf + m*C*dT1 = M*Lv + M*C*dT2

Using known values:

M = (m*Lf + m*C*dT1)/(Lv + C*dT2)

M = (0.425*3.33*10^5 + 0.425*4186*88)/(2.256*10^6 + 4186*12)

M = 0.129 kg = 129 gm

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