You have____kg of water at 273.2500k. The specific heat of water is 4186 J/Kkg.

Question

You have____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___g of ice at273.1500K. How much heat (in Joules) does it take to melt all that ice? Show me how you find how much ice (in grams) is unmelted 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)

heat to be removed Q1 = mwater*C*dt = 60*4186*(273.25-273.15) = 25116 J

(b)

heat required Q2 = mice*L = 0.08*333*10^3 = 26640 J

(c)

heat required to melt the ice is greater then the heat requred to cool the water

all the ice will not melt.eater and ice are equilibrium at 273.15 K

amount of ice remainig = m

haet removed = heat gained

mwater*Cw*dt1 = (mice-m)*L

mwater*Cw*dt1 = mice*L - mL

Q1 = Q2 - m*L

25116 = 26640 - (m*333*10^3)

m = 4.6 grams of ice

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