# Determining time to melt an ice cube

1. Jul 14, 2011

### mom2maxncoop

1. The problem statement, all variables and given/known data

The mass of the piece of ice is 0.25kg.
Heated by an electric heater (assume there is no loss of energy to the surroundings)
Started at -30°C at 0seconds
After 150s the ice cube is -10°C

2. Relevant equations
Q=mL$_{f}$
$\Delta$t=Q$/$P

3. The attempt at a solution
Q=mL$_{f}$
=(0.25kg)(330000J/(kg·°C)
=82500J

I know the next step is to use the second equation, but since I don't know the W of the heater, how would I go abouts finding out the time??

2. Jul 14, 2011

### Pi-Bond

How about using the time given (150 s for going from -30°C to -10°C) to find the power?

3. Jul 14, 2011

### mom2maxncoop

I've tried many different formulas and most don't make sense to me :/

For example this is one of the formulas I've gotten, but it just seems so wrong to be correct because once you put it into the other equation you get like 3 seconds which makes NO sense what so ever.
$\Delta$t=Q/P

150s = (-30°C-(-10°c)/P
150s = (20°C)/P
150s/20°C = P
7.5W= P

4. Jul 14, 2011

### Pi-Bond

You are making the wrong substitution for Q; you are substituting the temperature difference whilst you have to substitute mcΔT i.e. the heat Q required to change the temperature by ΔT. Try that out and see if it makes more sense.

5. Jul 14, 2011

### mom2maxncoop

Q=mc$\Delta\Delta$t
Q= (0.25)(2100)(0--30°C)
Q = 6300J

(now find the Power)
P=Q/$\Delta$t
P=6300J/150s
P=43W

Now I can use the first formula I figured out:
Q=mL$_{f}$
=(0.25kg)(330000J/(kg·°C)
=82500J

Next step would be to find the time!
$\Delta$time=Q/P
$\Delta$t=82500J/43W
=1918.65

Woo, I think it's all correct now! Thank you for your help. I knew what formulas to use after I was given a hint after the first one.

Again thank you!!
=1920s