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(11%) Problem 2: Suppose you place 0.27 kg of 22°C water in a 0.525 kg aluminum

ID: 1731664 • Letter: #

Question

(11%) Problem 2: Suppose you place 0.27 kg of 22°C water in a 0.525 kg aluminum pan with a temperature of 149°C, and 0.0125 kg of the water evaporates immediately, leaving the remainder to come to a common temperature with the pan Specific heat (c) Substances Solids J/kg C Aluminum Concrete Copper Granite Ice (average 900 840 387 840 2090 kcal/kg.C 0.215 0.20 0.0924 0.20 0.50 Liquids Water 4186 1.000 Gases 1520 (2020) 0.36300.482 ©theexpertta.com Steam (100°C) What would be the final temperature, in degrees Celsius, of the pan and water? The heat of vaporization of water is LV-2256 kJ/kg. You may neglect the effects of the surroundings and the heat required to raise the temperature of the vaporized water Grade Summary Deductions Potential Te = 122.11 0% 100% sin() cos() tan() ? ( cotanasin)acosO atan acotan) sinh) cosh tanhcotanh0 789 HOME Submissions A 456

Explanation / Answer

Lets say final temp is T .

Assuming no heat loss,

Q_released = Q_absorbed

Q_absorbed = (0.0125 x 2256 x 10^3) + (0.0125 x 4186 x (100 - 22)) + ((0.27 - 0.0125)(4186)(T - 22))

= 32281.35 + 1077.895 T - 23713.69

= 8567.6 + 1077.895 T

Q_released = 0.525 x 900 x (149 - T)

= 70402.5 - 472.5 T

So 8567.6 + 1077.895 T = 70402.5 - 472.5 T

T = 40 deg C ....Ans