Question

You are camping in a tent, you have some very hot water (700C) in a water bladder, but you realize that in fact you do not have enough drinking water. Your plan is to melt the block of ice you have seen in front of your tent using the boiling water you have and then treat the entire quantity. You estimate that the ice block has about 1 kg and is at a temperature of about -200C. How much water do you need to melt the ice and rise its temperature to 50C?

Answer 1

Using Energy conservation:

Heat gained by ice = Heat released by steam

Q1 + Q2 + Q3 = Q4 + Q5 + Q6

Q1 = Mi*Ci*dT1 = energy required to change -200.0 C ice into water

Q2 = Mi*Lf = energy required to change ice into water

Q3 = Mi*Cw*dT2 = energy required to change 0 C water into 50 C water

Q4 = Ms*Cw*dT3 = energy required to change 100 C water into 50 C water

Q5 = Ms*Lv = energy required to change steam into water

Q6 = Ms*Cs*dT4 = energy required to change 700 C steam into 100 C steam

Mi*Ci*dT1 + Mi*Lf + Mi*Cw*dT2 = Ms*Cw*dT3 + Ms*Lv + Ms*Cs*dT4

Mi = Mass of ice = 1 kg & Ms = mass of steam

from above expression

Mi*(Ci*dT1 + Lf + Cw*dT2) = Ms*(Cw*dT3 + Lv + Cs*dT4)

Ms = Mi*(Ci*dT1 + Lf + Cw*dT2)/(Cw*dT3 + Lv + Cs*dT4)

Now known values are:

dT1 = 0 - (-200) = 200, dT2 = 50 - 0 = 50 C, dT3 = 100 - 50 = 50 C, dT4 = 700 - 100 = 600 C

Ci = Specific heat of ice = 2090 J-kg/C

Cw = Specific heat of water = 4186 J-kg/C

Cs = Specific heat of steam = 2020 J-kg-C

Lf = latent heat of fusion = 3.34*10^5 J/kg

Lv = latent heat of vaporization = 2.25*10^6 J/kg

Using these values:

Ms = 1*(2090*200 + 3.34*10^5 + 4186*50)/(4186*50 + 2.25*10^6 + 2020*600)

Ms = 0.262 kg = required amount of hot water

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