Question

Two trials were performed in an experiment to determine the latent heat of vaporization of water at 100°C. The values of LV of water obtained were LV1 = 336 cal/g and LV2 = 338 cal/g. Find the percent difference between the two values. Find the percent error for each measurement if the accepted value of LV of water at 100°C is 339 cal/g.

Answer 1

Answer:

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Answer 2

Answer:

Percent error in each case is 0.88% and 0.29%  respectively

Explanation:

The latent heat of vaporization of water measured first LV1= 336 cal/g

The latent heat of vaporization of water measured second LV2=338 cal/g

accepted value of Latent heat of vaporization of water is =339cal/g

error in LV1 = |339-336|=3

error in LV2=|339-338|=1

Percentage error for LV1 =( error in LV1/accepted value )*100 =$\frac{3}{339}*100=0.88 %$%.

Percentage error for LV2 =( error in LV2/accepted value )*100=$\frac{1}{339}*100=0.29$%.