Substances Specific heat c (14%) Problem 5: A 0.035-kg ice cube at 30.0°C is pla
ID: 1609619 • Letter: S
Question
Substances Specific heat c (14%) Problem 5: A 0.035-kg ice cube at 30.0°C is placed in 0.315 kg of 35.0°C Solids JIkg C kcal/kg. water in a very well-insulated container. The latent heat of fusion for water is Lf 79.8 kcal/kg. Aluminum 8400 020 Concrete Copper 387 0.0924 Glass 840 Ice (average) 2090 0.50 1.000 Water 4186 Gases Steam (100 C) |1520 (2020) 0.363 (0.482) Otheexpertta.com What is the final temperature of the water, in degrees Celsius? Grade Summary Deductions 0% ial 100% Submission Attempts remaining: S per attempt) cotano asin0 acoso detailed view atan0 acotano sinh0 ODegrees O Radians Submit Feedback: deduction per feedback. Hints: deduction per hint. Hints remaining LExplanation / Answer
Amount of Heat lost by water due to change in its temperature,
Qlost = mw*cw*(T - Tw)
where T = final temperature of water
Amount of Heat gain by ice due to change in its temperature,
Qgain1 = -mi*ci*Ti + mi*cw*T
Amount of Heat gain by ice due to change in its phase from ice to water,
Qgain2 = mi*Li
where Li = latent heat of fusion of ice
At equilibrium temperature,
Qlost = -(Qgain1 + Qgain2)
mw*cw*(T - Tw) = mi*ci*Ti - mi*cw*T - mi*Li
By rearrangening the above equation,
T = mi*ci*Ti - mi*Li + mw*cw*Tw / (mw*cw + mi*cw)
T = 0.035*0.5*(-30) - 0.035*79.8 + 0.315*1*35 / (0.315*1 + 0.035*1)
T = 22.02 oC