How Long Does a 2.02 kW Heater Take to Match the Heat from Solar-Heated Water?

AI Thread Summary
To determine how long a 2.02 kW heater would take to match the heat from 859 kg of solar-heated water cooling from 11.3 to 0 °C and freezing, the total heat (Q) required is calculated as 328,406,952 J. The heater's power output is 2.02 kW, equivalent to 2020 J/sec. Dividing the total heat by the heater's output gives the time required: 328,406,952 J / 2020 J/sec, resulting in approximately 162,000 seconds or about 45 hours. The initial calculation mistakenly used a 1 kW heater instead of the correct 2.02 kW. The correct operation time for the 2.02 kW heater is significantly less than the initially calculated 91 hours.
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Homework Statement


To help keep his barn warm on cold days, a farmer stores 859 kg of solar-heated water (Lf = 3.35E+5 J/kg) in barrels. For how many hours would a 2.02 kW electric space heater have to operate to provide the same amount of heat as the water does, when it cools from 11.3 to 0 °C and completely freezes?

Homework Equations


Q=mc(deltaTdegC)+mL

The Attempt at a Solution


Q=(859)(4187)(11.3)+(859)(335000)
Q=328406952 J
1kW=1000W=1000J/sec
1000t=328406952 J
t=328406 secs*1min/60secs*1hour/60mins=91 hours

But this is wrong. What am I missing?
 
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You have calculated for a 1000w heater.
 
You're correct that 1 kW = 1000 J/sec. But the heater is a 2.02 kW heater, not a 1 kW heater.

[Edit: like NascentOxygen said.]
 

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