What is the change in internal energy of the water

[Kristmas]

Hi there, I have some serious problems with a couple of questions. Need help for exam revision.
1.1 Homework Statement
A 2kg chunck of ice at -10Celsius is added to 5kg of water at 45Celsius. What is the final temperature of the system?

1.2 Relevant equations
specific heat capacity of ice - 2100 Jkg^-1K^-1
specific heat capacity of water - 4190 Jkg^-1K^-1
latent heat of fusion for ice - 334 000 Jkg^-11.3 The attempt at a solution
Using formula E=mC(change in)T
E₁ - 2 x 2100 x 10 = 42000J
E₂ - 2 x 334 000 = 668 000J
E₃ - 668 000 + 42 000 = 710 000J
E₄ - 710 000=5 x 4190 x (change in)T T=33.89Celsius, so final temperature is 45-33.89=11.11, but the correct answer has to be 7.9Celsius.

2.1 Homework Statement
A cylinder with a piston contains 0.2 kg of water at 100 Celsius. What is the change in internal energy of the water when it is converted to steam at 100Celsius at a constant pressure of 10⁵ Nm^-2?

2.2 Relevant equations
latent heat of vaporisation of water - 2.26 x 10⁶ Jkg^-1
density of water - 1000 kgm^-3
density of steam - 0.6 kgm^-3

2.3 The attempt at a solution
Using (change in) U=(change in)Q - P x (change in)V
U=internal energy, Q=heat, P=pressure, V=volume
Change in heat is 2.26 x 10⁶ x 0.2 = 452 000 J
Can't manage more :(.
Supposed answer is 418.7 kJ.

 

[ehild]

"E4 - 710 000=5 x 4190 x (change in)T"

This is not correct.

You have to warm up to the 2kg 0 oC melted ice to the final temperature.

 

[Kristmas]

Sorry, I can't see it still. I know it has taken 710 000Joules of energy to get the ice to 0 Celisu and melt completely to water. So I thought i simply take that energy away from the water at 45 Celsius. Thus using Energy=mass*specific heat capacity of water* change in temperature to get the change in temperature from 45.

 

[ehild]

It is true that you need that 710000 J to warm up and melt the ice to oC water. That energy is taken away from the 45 oC water, but you still have 2 kg 0 oC water to be warm up to the final temperature.

ehild

 

[Kristmas]

Thank you Ehil, I got the answer to question 1 now.
Worked out solution to question 2 as well.
Change in heat is 0.2 x 2.26 x 10⁶ Jkg^-1=452 000 Joules
Pressure is already given at 100 000 Pa
Original volume is mass/density, so 0.2/1000=0.0002m³.
Mass stays the same, so 0.2kg of steam occupies: 0.2/0.6=0.333333 m3 of space. Change in volume is the volume of steam - volume of water, 0.333333-0.0002=0.333133m³.
So using the overall equation: 452 0000J - 100 000Pa x 0.333133m³=418 686.7 J.
Can close the thread.