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(10%) Problem 4: You are on your way to a party when the host asks you to pick up a bag of ice. At the grocery store you grab a 7. kg bag that was kept at a temperature of -3.7 C. When you get to the party, you find a large cooler to put the ice in. There is already 28 L (ie., 28 kg) of water in the cooler at a temperature of 21°C. You toss the ice into the water and close the lid. The specific heat and latent heat of fusion for water are 4186 J/(kg C) and 3.34 x 105 Jikg, respectively. The specific heat of ice near its freezing point is 2000 J/(kg C). or Find the temperature, in degrees Celsius, of the water in the cooler after the party. Assume the ice maintains its its on the way to the party and the cooler is well insulated. Grade Summary npleted mpleted Attempts remaining: 992 cotan0 in acos0 detailed view atan0 acotan0 sin 0 mpleted l give u Hint Feed
Find the temperature, in degrees Celsius, of the water in the cooler after the party and the cooler is well insulated. on the way to the Grade Summary Deductions Potential 100% sin0 cotan0 asin0acos0 atan0 acotan0 sinh01 Attempts remaining: 997 (0% per attempt detailed view 2 3 0END Submit Hint I give up! Feedback: % deduction per feedback. 31 MacBook Ain
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Answer #1

Mass of ice = m1 = 7 kg

Mass of water = m2 = 28 kg

Initial temperature of the ice = T1 = -3.7 oC

Initial temperature of the water = T2 = 21 oC

Melting point of ice = T3 = 0 oC

Final temperature of the mixture = T4

Specific heat of ice = Ci = 2000 J/(kg.oC)

Specific heat of water = Cw = 4186 J/(kg.oC)

Latent heat of fusion of water = L = 334000 J/kg

The heat lost by the water is equal to the heat gained by the ice.

m1Ci(T3 - T1) + m1L + m1Cw(T4 - T3) = m2Cw(T2 - T4)

(7)(2000)(0 - (-3.7)) + (7)(334000) + (7)(4186)(T4 - 0) = (28)(4186)(21 - T4)

51800 + 2338000 + 29302T4 = 2461368 - 117208T4

146510T4 = 71568

T4 = 0.49 oC

Temperature of the water in the cooler after the party = 0.49 oC

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