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# specific heat capacity questions help watch

1. i'm a bit stuck on a few questions can any1 plz help?
a 3kw heater is used to turn boiling water to steam it is on for 5mins calculate the max mass of steam that can be produced. I did
P=energy/time and used Q=mL equation to get m to be 3.98 x 10^-4 Kg is this right? Also other than heat loss to surroundings why might the actual amount be lower than this?
2. (Original post by panjabiflower)
i'm a bit stuck on a few questions can any1 plz help?
a 3kw heater is used to turn boiling water to steam it is on for 5mins calculate the max mass of steam that can be produced. I did
P=energy/time and used Q=mL equation to get m to be 3.98 x 10^-4 Kg is this right? Also other than heat loss to surroundings why might the actual amount be lower than this?
Haven't check you number but sounds plasuable, I remember it is quite low. The actual amount might be lower as the heater might also heat the steam to above 100°C, so some energy used up.

Just ensure you converted minutes to seconds!

EDIT: Checked this and I got about 0.443 KG taking latent heat to be 2.23KJ/g or 2.23MJ/kg. So I think you might have used 5 minutes not 300s.
3. on the data sheet given by my teacher it said speciifc latent heat of vaporisation of water= 2260000Jkg^-1
I got the energy= 900000J
Q=mL
900000= m x 2260000
m= 0.398Kg (oops made a mistake in above post lol) is this right?
4. Also can there be heat loss within the kettle itself? i.e not only making the water boil to steam
- can internal energy be lost in the kettle?
5. I'm sure I might have an error in the calculation, but I got the same energy as you, did do it quite fast. You could include energy lost due to specific heat capacity of the element, and body of the kettle (although if water is already boiling then the kettle is likely to be at 100°C and little change in temperature). Also the kettle element may not deliver all the energy into heat, so lost in the kettle wiring etc.
6. when working out the specific heat capacity of a liquid how can errors be reduced?
7. (Original post by panjabiflower)
when working out the specific heat capacity of a liquid how can errors be reduced?
More insulation, accurate equipment, try to find specific heat capacity of equipment using a known substance like water, so you know the energy input into the water and temp it changed, so the lost energy will have gone into the container and air. So once you know this, the you can account of this when measuring an unknown substance.
8. 0.4kg aluminum pan contains 1.6kg water at 15 degrees when placed on a cooker so that the water is heated to 90 degrees what percentage of the energy supplied has gone to heating the pan?
9. (Original post by panjabiflower)
0.4kg aluminum pan contains 1.6kg water at 15 degrees when placed on a cooker so that the water is heated to 90 degrees what percentage of the energy supplied has gone to heating the pan?
I hope you are getting this now but I'll try to help: You have specific heat capacity of water (w) and aluminium (a). So do:

0.4*a*(90-15)
--------------------------------- x 100
(0.4*a*(90-15)) + (1.6*w*(90-15))

So this find what proportion of energy is used to heat the pan over total energy used. Then times by 100 to get the percentage.
You should get around about 1-8% at a guess, and specific heat of aluminium is quite a bit less of water, and there is also less mass.
10. i was just wondering why you wouldn't do 0.4*a*(90-15)/ (1.6*w*(90-15)
11. (Original post by panjabiflower)
i was just wondering why you wouldn't do 0.4*a*(90-15)/ (1.6*w*(90-15)
Because that is the ratio of energy because that would find the percentage of energy used to heat the pan out of the energy used the heat the water. The total energy input is shared between the pan and the water, so you need my equation above.

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