- #1
Kristmas
- 3
- 0
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.
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.