You haw have 100 kg of water at 273.2500K. The specific heat of water is 4186 J/

Question

You haw have 100 kg of water at 273.2500K. The specific heat of water is 4186 J/Kkg. How much heat (in Joules) must be removed from the water to cool it all down to 273.1500K? Water has a latent heat of fusion of 333 kJ/kg. You have 90 g of ice at 273.1500K. How much heat (in Joules) does it take to melt all that ice? Show me how you find the final temperature (in mK) when the water and the ice combine. You must show your work. This problem is your opportunity to show me the physics concepts you know, take advantage of it, and include units.

Explanation / Answer

25. a)

In this problem, the heat calculation involves a temperature change:

Q = Mw*cw*(Tf - Ti)   ;    Q = 100 kg * 4186 J/Kkg * 0.1 K = 41860 J  

b) Q=333 kJ/kg*0.09 kg=29.97kJ

c) Latent heat to ice at 0°c to convert into saturated water at 0°c + sensible heating of this saturated water from 0°c to final temp T°c = cooling of hot water from 80°c to final temp T°c

=> (100*333kJ) + {100*1*(T-0)} =
=> 100*T = -100*T

=> 200*T = 0

=> T = 0

So final temp of mixture is 0°c and mixture is saturated water.

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