Time to Heat 125L Water to 60C

[GoodOldLimbo]

Homework Statement

A tank contains 125 litres of water. If it's temperature is 20C, how long would it take for a 2kW immersion heater to raise its temperature to 60C if the average rate of heat loss is 150W?

Homework Equations

I am using the specific heat capacity equation:

ΔQ = mcΔT

The Attempt at a Solution

ΔQ = 125 x 4200 x 40 = 21, 000, 000 J

2kW = 2kJ per second.

Concerning Heat Loss:
2000W - 150 W = 1850 W = 1.85kW

21,000KJ = 11, 351 seconds.

11, 351 / 60 = Approximately 190 minutes.

I kind of get the feeling I'm not right. Any tips? Would be appreciated.

 

[Simon Bridge]

Unless your course includes thermal mixing and nonlinear differential equations, that seems to be fine.
The way you could check this sort of thing is relate it to common experience ...

Your figure works out to 1.5mins to heat 1L of water by 40deg, or 3mins to start it boiling.
A good-sized cup would be about 1/3L ... or 1min to start boiling.
Does that seem reasonable?

An electric kettle is usually 2-3kW ... does it normally take about a minute or so to boil a cup of water?

 

[dauto]

Homework Statement

A tank contains 125 litres of water. If it's temperature is 20C, how long would it take for a 2kW immersion heater to raise its temperature to 60C if the average rate of heat loss is 150W?

Homework Equations

I am using the specific heat capacity equation:

ΔQ = mcΔT

The Attempt at a Solution

ΔQ = 125 x 4200 x 40 = 21, 000, 000 J

2kW = 2kJ per second.

Concerning Heat Loss:
2000W - 150 W = 1850 W = 1.85kW

21,000KJ = 11, 351 seconds.

11, 351 / 60 = Approximately 190 minutes.

I kind of get the feeling I'm not right. Any tips? Would be appreciated.

Your answer seems fine. I'd just take marks down for the statement "21,000KJ = 11, 351 seconds." which is obviously wrong.

 

[Simon Bridge]

dauto is correct - someone marking what you wrote will expect you to be articulate.
It is easy to see what you mean, but it is sloppy notation.

 

[HalfLostAndSearching]

Your approach is correct, however, there are a few things to consider. Firstly, the specific heat capacity of water is 4,200 J/kg°C, not 4,200 J/L°C. So, the mass of water should be converted from liters to kilograms. This can be done by multiplying 125 L by the density of water, which is approximately 1 kg/L. This gives a mass of 125 kg.

Secondly, the heat loss rate should be calculated as a percentage of the heater's power. In this case, it would be (150/2,000) x 100 = 7.5%. This means that only 92.5% of the heater's power is actually heating the water.

With these corrections, the calculation would be as follows:

ΔQ = 125 x 4,200 x 40 = 21,000,000 J

2kW = 2,000 J/s

11,351 seconds x (1/0.925) = 12,267 seconds or approximately 205 minutes.

So, the estimated time to heat 125 L of water from 20°C to 60°C with a 2kW heater and a heat loss rate of 150W would be approximately 205 minutes. However, this is just an estimate and the actual time may vary depending on other factors such as the efficiency of the heater and the insulation of the tank.