Solve Energy Questions: 6 & 7 | Hiya Cathy

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In summary: W kettle in one minute and the other involving the specific latent heat of vaporisation of water. In summary, the correct answers for the first question are 66g and 66ml/66cm^3, and for the second question, the specific latent heat of vaporisation can be calculated by dividing the heat input during one minute by the heat of vaporization.
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CathyLou
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Hiya.

I'm really stuck on these two questions and I've no idea how to really approach them at all. Any help would be really highly appreciated.

6. How much water does a 2.5 kW kettle boil off in one minute, in terms of

(a) mass, and
(b) volume of water?

For the above question my original working was:

(a) 2.5kW x 60s = 150kJ generated per minute.

Change in energy = mass x specific heat capacity x change in temperature

Assuming that the water starts at 25ºC, we need to raise the temperature by 75ºC to get to 100ºC.

150000 = m x 4200 x 75

m = 0.48 kg

m = 476g

(b) 476g = 476ml or 476 cm^3

I know that these answers are wrong as yesterday in class the teacher said that the correct answers are 66g and 66ml/66cm^3.


7. An electric heater immersed in some water raises the temperature from 40ºC to 100ºC in 6 min. After another 25 min it is noticed that half the water has boiled away. What value does this experiment give for the specific latent heat of vaporisation of water?

All that I wrote down for the next one was:

Change in energy = mass x the latent heat of vaporisation

Change in energy = mass x specific heat capacity x change in temperature

Mass = density x volume


Thank you very much.

Cathy
 
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Thanks for your help. I've got the correct answer now. :smile:

Cathy
 

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