Part A: How much heat (in kJ) is required to warm 13.0 g of ice, initially at -1
ID: 475124 • Letter: P
Question
Part A: How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 C, to steam at 111.0 C? The heat capacity of ice is 2.09 J/gC and that of steam is 2.01 J/gC.
Part B An 8.5-g ice cube is placed into 255 g of water. Develop an equation that can be used to calculate the temperature change in the water upon the complete melting of the ice given an initial temperature of T. Assume that all of the energy required to melt the ice comes from the water. Express your answer in terms of the initial temperature of water, T.
Explanation / Answer
Q = Q1 + Q2 + Q3 + Q4 + Q5
Q1 = ice to water at 0C
Q1 = m*Cs*dT = 13*2.09*[10-0] = 271.7 J
Q2 = ice at 0C to water at 0C
Heat of fusion = is 6.02 kJ/mol
18 gms ----> 6.02 KJ
13 gms = 4.347 KJ
Q3 = Water at 0C to Water at 100 C
Cliq = 4.08
Q3 = m*Cliq*dT = 13*4.08*[0-100] = 5304 J
Q4 = Water at 100 C to Vapour at 100 C
Q4 = Latent heat of vapourization = 40.65 kJ/mol
13 gms = 40.65*18/13 = 56.28 KJ
Q5 = Steam at 100C to 111C
Q5 = m*Csteam*dT
Q5 = 13*2.01*[100-111] = 287.43 J
Q = Q1+Q2+Q3+Q4+Q5 = 66490.14 J = 66.49 KJ
Part B :
You must consider the the heat exchange. You have the latent heat when the ice becomes liquid.
The latent heat is Ql = cl*m with m = 8.5 gram = 0.0085 kg and the latent heat constant in fusion cl = 335 J/gram so Ql = 335*8.5 = 2847.5 J.
This latent heat is absorbed to the water so the water's temperature goes down of T. The heat constant of liquid water is cp = 4186 J/kg*K so:
Ql = ma*cp*T with ma = 0.255 kg (255 grams) so:
T = Ql/(ma*cp) = 2847/(4186*0.255) = 2.68°C.