physics question?

Question

To help keep his barn warm on cold days, a farmer stores 890 kg of solar-heated water (Lf = 3.35 105 J/kg) in barrels. For how many hours would a 1.0-kW electric space heater have to operate to provide the same amount of heat as the water does, when it cools from 10.0 to 0.0°C and completely freezes?
__________ hours


i am really confused what to do. i thought that the 1.0 kW was work, which equals q/t. so i solved for q by adding the q that it would take to get to zero and the q for fusion. BUT when i went to solve for t i got a negative answer. so i know it must be wrong. and redid it and got a huge number. any advice on how to take this problem?

Answer 1

Joules are units or energy, which is work.
Watts are units of power, which is work/time.

There are two phases to consider with the water in terms of energy transfer:

The energy to cool the water from 10 to 0, and the heat of fusion to freeze the water.

The specific heat of pure water water is
1 calorie/gram °C = 4.186 joule/gram °C
your problem statement indicates the Lf as 3.35 105 J/kg
while I believe that Lf for water is 334 kJ/kg
is that what you meant (334 vs 335 is no big deal)?

So, the first energy transfer is the cooling of the water from 10 to 0 C
I will work in kJ/kg
10*4.186*890 kJ
37255.4 kJ
The energy transfer during freezing is

335*890 kJ
298150 kJ
adding them together to get the total work
335405.4


Now convert to kW-hr

a 1 kW heater transfers 1 kJ/sec
there are 3600 seconds in an hour
so the total time equivalent is

335405.4/3600
93.2 hours
j

Answer 2

AP Physics is tough. If your math grades are at least a B then I would consider taking AP Physics. It will be tough for you unless you are really strong in math.