Suppose you place 0.255 kg of 22.5 degree C water in a 0.525 kg aluminum pan wit

Question

Suppose you place 0.255 kg of 22.5 degree C water in a 0.525 kg aluminum pan with a temperature of 153.5 degree C, and 0.0115 kg of the water evaporates immediately, leaving the remainder to come to a common temperature with the plan. What would be the final temperature, in degree Celsius, of the pan and water. The heat of vaporization of water is L = 2256 KJ/kg. you my neglect the effects of the surroundings and the heat required to raise the temperature of the vaporized water. Numeric : A numeric value is expected and not an expression.

Explanation / Answer

Heat required for evaporation = m*C*delta T + m*Lv
= 0.0115*4186*(100-22.5) + 0.0115*2256000
= 29674.7725 J

remaning water = 0.255 Kg - 0.0115 Kg = 0.2435 Kg

Let final temperature be ToC

use:
Heat required = heat given
29674.7725 + 0.2435*4186*(T-22.5) = 0.525*900*(153.5-T)
29674.7725 + 1019.291 T - 22934.0475 = 72528.75-472.5*T
T= 44.1 oC
Answer: 44.1 oC

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