The details of the experiment is given partially.
It requires information of energy imput given to the system . We suppose it depends on the time , and you calculated correctly, and the energy is E say. (but you did not mention what is E.
The quantum of heat utilsed by 1000 g of water could easily calculated if the initial temerature of the water before heating and at the boiling point are given. The energy is the product of mass of water and the difference temperature in Kelvin and 4.186/1000 K Joule.
Make an allowance for the kettle's energy raise as it also raises in temperature.
So the efficiency of the inter conversion is = {Energy due to raise in temperature of kettle+energy utilised by 1000 g water to raise its temperature to boiling point)/supplied enrgy}*100 %.
Therefore the energy loss = (100-efficiency in percent)%
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