Question

In: Physics

An insulated container contains 15.0 g of steam at 100C. A mass of 65.0 g of...

An insulated container contains 15.0 g of steam at 100C. A mass of 65.0 g of ice at 0.00C is dropped into the container (assume no energy is absorbed by the container).

C = Celsius

a). How much heat is released by the steam when it condenses at 100C? (Qs)

b). How much heat is absorbed by the ice when it melts at 0.00C? (Qi)

c). What is the final temperature of the the water in the container? (T)

Expert Solution

Part A.

Energy released by steam when it is condensed will be:

Q1 = Ms*Lv

Lv = latent heat of vaporization = 2.26*10^6 J/kg

Ms = mass of steam = 15.0 g = 15.0*10^-3 kg

Q1 = 15.0*10^-3*2.26*10^6 = 33900 J

Part B.

Energy released by ice when it is melted will be:

Q2 = Mi*Lf

Lf = latent heat of fusion = 3.34*10^5 J/kg

Mi = mass of ice = 65.0 g = 65.0*10^-3 kg

Q2 = 65.0*10^-3*3.34*10^5 = 21710 J

Part C.

Suppose final temperature of water is T, then

energy absorbed by ice = energy released by steam

Q2 + Q3 = Q1 + Q4

Q3 = energy absorbed by water at 0 C to T C = Mi*Cw*dT3

Q4 = energy released by water at 100 C to T C = Ms*Cw*dT4

dT3 = T - 0 = T

dT4 = 100 - T

Cw = Specific heat capacity of water = 4186 J/kg

So,

21710 + Mi*Cw*dT3 = 33900 + Ms*Cw*dT4

65.0*10^-3*4186*T = 12190 + 15.0*10^-3*4186*(100 - T)

T = [12190 + 15.0*10^-3*4186*100]/(80*10^-3*4186)

T = 55.15 C = final temperature

Please Upvote.

genius_generous answered 8 months ago

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