# Help with 2 questions quickly for exams please

• Conor11
In summary, the conversation discusses the effects of water evaporation on the temperature of air in a kitchen and in a refrigerator. It provides equations for calculating the rate of energy transfer from the water by evaporation, the rise in temperature of the air, and the energy required to cool the air in the refrigerator. It also mentions the use of specific heat capacity and specific latent heat of vaporization in these calculations.
Conor11

## Homework Statement

. A saucepan containing water is left boiling on a cooker in a kitchen for some time. As the water evaporates from the saucepan and then condenses , the temperature of the air in the kitchen is raised. Given that the volume of air in the kitchen is 15.6m^3 and that water is evapoarting from the saucepan at a rate of 0.65 grams per second calculate. 1. the rate at which energy is transferred from the water in the saucepan by evaporation. 2. the rise in temperature of the air caused by the evaporation ad condensation after 5 mins , assuming that 10% of the transferrred energy is absorbed by the air. (specific heat capacity of air= 1.0 e^3 jk^-1k-1, density of air = 1.2kg m-3, specific latent heat of vaporisation of water 2.3 e^6 j kg-1)

2. Given that the specific heat capacity of air is 1.0 e^3 jkg-1k-1 calculate how much energy must be removed from the refridgerator conatining 0.15m^3 of air to lower the temperature of the air from 20 C to 4 C . (density of air=1.2 kg m-3) The air in the refridgerator is cooled by allowing a liquid to evaporate in a closed pipe inside the refridgerator. The vapour is the pumped through the pipe to the outside of the fridge, where it condenses again. Given that 2.3 grams of the liquid evaporate per minute how long will it take to cool the air in the fridge from 20 C to 4 C if the specif laten heat of vaporisation of the liquid is 2.5 e5 j Kg-1? (heat loss by fridge is negligble)
21 hours ago - 3 days left to answer.

## The Attempt at a Solution

1. 0.006592.3e60= 1495
not sure what to do at all next , the time it takes and the density have to be taken into account but i don't see how.

2. m=dv
therefor m=0.18
e=mcdelta t
therefore e=2880j
Dont understand how to calculate using the specific latent formula with the rest of the question.

The energy to vaporize 1 g of water is the specific latent heat of vapor...
lets call it $E_{vap}$
You know the mass per time and that the above energy is in the unit J/g or J/kg:
$\Delta E = E_{vap}m$
2. now you have the energy $\Delta E$ and you know the specific heat capacity of air, so multiply the energy found in 1 by the time and use. And remember that only 10 % goes to the air:
$0.10\Delta E t = m C \Delta T=\rho V C \Delta T$
Do you get it?

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