150 g of water and 50.0 g of ice are in a container made from 20.0 g of aluminum

Question

150 g of water and 50.0 g of ice are in a container made from 20.0 g of aluminum. The system consisting of the water, ice and container are at thermal equilibrium at 0degreeC. The container is wrapped in a thermal insulation that isolates it from the environment. A water warmer rated at 200 watts is placed in the container and switched on. The contents of the container are gently stirred to keep the temperature of the system uniform. Determine the amount of heat (in joules) required to raise the temperature of this system to 100degreeC. Lf = 333 J/g, cw = 4.186 J/(g.K). cA1 = 0.900 J/(g.K). How long will it take the water warmer to heat the mixture to 100degreeC? Show calculation. After the system has been heated to 100degreeC, an additional 50.0 g of ice at 0degreeC are added. What is the final temperature of the mixture? Show work.

Explanation / Answer

a)

heat required to raise the temperature of this system to 100C = 50*333 + 200*4.186*(100 + 273) + 20*0.9*(100 + 273) = 335639.6 J

b)

it will take t sec.

power = Q/t

t = 335639.6/200 = 1678.198 sec = 27.97 min

c)

heat lose = heat gain

200*4.186*(100+273 - T) + 20*0.9*(100+273 - T) = 50*333 + 50*4.186*T

312275.6 - 837.2*T + 6714 - 18*T = 16650 + 209.3*T

335639.6 = 1064.5*T

T = 315.3 K = 42.3 0C

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