Calculating the Ampere Rating of a Circuit Breaker

[dewdrop714]

The problem statement:
To save on heating costs, the owner of a greenhouse keeps 800kg of water around in barrels. During a winter day, the water is heated by the sun to 50 degrees Fahrenheit. During the night the water freezes into ice at 32 degrees Fahrenheit in 10 hours. An electrical heating system is used providing the same heating effect as the water.


The attempt at a solution:

P=Q/t = (542*10^6)/36000s = 15066 Watts
I=P/V = 15066/240 = 62.77Amperes

The minimum ampere rating that the 240 volt circuit breaker would have to be to avoid tripping is what i calculated...it turned out to be 62.77Amperes.

The question:
What I don't get is "does the circuit breaker trip? Why or why not?"

 

[LowlyPion]

How did you arrive at the 5.42*10⁶ J for Q?

 

[dewdrop714]

Well 4186 is spec heat of water and 2093 is spef heat of ice. I also converted the temps to kelvins. Then I input them into the equations...I did Q1= (800)(4186)(305.93)=102*10^7 and Q2=(800)(2093)(287.93)=482*10^6. Then i got Q= Q1-Q2 = 542*10^6.

 

[LowlyPion]

First of all I would just convert to °C making 50°F = 10°C

Next I would point out that what you have is a Specific heat of water over 10°C + the heat of fusion of water, not the specific heat of ice as there is no change in ice temp below 0°.

http://230nsc1.phy-astr.gsu.edu/hbase/thermo/phase.html#c1

That would be :

Q = (ΔC*4186j/°Ckg + 334*10³J/kg)*800kg
Q = (10*4186 + 334,000)*800 = 3.7586*10⁵*800 = 3*10⁸