An earthenware vessel loses 1 g of water per second due to evaporation. The water equivalent of the vessel is 0.5 kg and the vessel contains 9.5 kg of water find the time required for the water in the vessel to cool to `28^@C` from `30^@C`. Neglect radiation losses. Latent heat of vapourization of water in this range of temperature is `540 cal//g`.

Options

A. 38.6 min
B. 30.5 min
C. 34.5 min
D. 41.2 min

Answer

A

Explanation

As water equivalent of pitcher is 0.5 kg i.e.k pitcher is equivalent to 0.5 kg of water, heat to be extracted from the system of water and pitcher for decreasing its temperature from 30 to `28^@C` is
`Q_1=(m+M)cDeltaT`
`=(9.5+0.5)kg((1kcal)/(kgC^@))(30-28)^@C`
And het extractged from the pitcher through evaporation in t minutes
`Q_2=mL=[(dm)/(dt)xxt]L=[(1g)/(min)xxt]580 cal//g`
According to given problem `Q_2=Q_1`, i.e., `580xxt=20xx10^3`
`t=34.5min`