- #1
mertzi
- 12
- 0
Member warned about lack of template
Hi! I'm new here and I can't find anything helpful through google so I thought I'd give PF a try. Sorry for not using the template fully but I have no ideas on how to solve this problem.
1. 0.1kg ice at 263K is added to 1kg water at 273K. Calculate new mass of the ice.
2. I know how to do this calculation if the water temperature is higher than 273K, but then I got this problem where the water is already at 273K (0°C) therefore no change in temperature so I can't use Q=c⋅m⋅ΔT or at least I can't get my brain around how. Is there an "opposite direction" enthalpy formula? As I have understood it this situation will actually increase the mass of the ice since energy will be released when the ice is heated to 273K.
3. Can I do this the other way around i.e. calculate how much water needs to turn to ice to heat the ice to 273K?
ΔQ=cice⋅mice⋅ΔT = 2.2kJ/(kg⋅K)⋅0.1kg⋅10K = 2.2kJ
Then
2.2kJ = m⋅Lwater
m = 2.2/332 ≈ 6.6⋅10-3kg
mice total = mice 0+mice 2 = 0.1+6.6⋅10-3 = 0.1066kg
Thanks!
1. 0.1kg ice at 263K is added to 1kg water at 273K. Calculate new mass of the ice.
2. I know how to do this calculation if the water temperature is higher than 273K, but then I got this problem where the water is already at 273K (0°C) therefore no change in temperature so I can't use Q=c⋅m⋅ΔT or at least I can't get my brain around how. Is there an "opposite direction" enthalpy formula? As I have understood it this situation will actually increase the mass of the ice since energy will be released when the ice is heated to 273K.
3. Can I do this the other way around i.e. calculate how much water needs to turn to ice to heat the ice to 273K?
ΔQ=cice⋅mice⋅ΔT = 2.2kJ/(kg⋅K)⋅0.1kg⋅10K = 2.2kJ
Then
2.2kJ = m⋅Lwater
m = 2.2/332 ≈ 6.6⋅10-3kg
mice total = mice 0+mice 2 = 0.1+6.6⋅10-3 = 0.1066kg
Thanks!
Last edited: