Physics, Temperature Problem?

Question

To help keep his barn warm on cold days, a farmer stores 840kg of solar-heated water (Lf = 3.35X10E5 J/kg) in barrels. For how many hours would a 2.0k-W electric space heater have to operate to provide the same amount of heat as the water does when it cools from 10.0 celsius to 0 degrees celsius and completely freezes?

Whats the answer? Can anyone help? Thanks in advance!

Answer 1

Specific heat capacity of water = 4200 J kg^-1 K^-1

Heat lost by water = 840 X 4200 X (10.0 - 0)

= 35 280 000J

Heat lost during freezing = 840 X 3.35 X 10^5

= 281 400 000J

Total heat lost = 35 280 000 + 281 400 000

= 316 680 000J

Power X time = heat energy

time = 316 680 000 / 2000  

= 158 340 s

= 158 340 / 3600

= 43.98 hours (ans)

Answer 2

The energy released by cooling water is one calorie per gram per degree C of cooling.  As each gram of water loses 10 degrees of temperature, it releases 10 calories from this cooling.  Each gram of water also loses a further 80 calories of heat as the water freezes.  This is a total of 90 calories of heat per gram.  Multiplied by the conversion factor of 4.182 joules per calorie, this means each gram of water releases about 376 joules of heat as it cools and freezes.  Multiplied by 8.40e5 grams, that amounts to 3.16e8 joules of heat.

A kilowatt is a measure of power, which means rate of energy transformation.  One watt of output means a release of one joule of heat per second.  The time the heater must run to produce 3.16e8 joules is therefore 3.16e8 J / 2.0e3 W or 1.58e5 seconds or 43 hours 52 minutes.