# Latent Heat, Some Density, and God knows what else

1. Jan 27, 2012

### platachog

I've been having trouble understanding this second semester of physics thus far. I've done SHM pretty well, but this stuff is just strange to me. So here are the problems verbatum, but i'll fill out the template below to help ease the pains.

For problem 1:
1. The problem statement, all variables and given/known data
The heating element of a water heater in an apartment building has a maximum power output of 27 kW. Four residents of the building take showers at the same time, and each receives heated water at a volume flow rate of 13x10-5 m3/s. If the water going into the heater has a temperature of 10°C, what is the maximum possible temperature of the hot water that each showering resident receives?

P = 27 kW
Volume Flow Rate = 13E-5 m^3/s
T(i) = 10°C
T(f) = ?

2. Relevant equations
Q = mcΔT
ρ = m/v

3. The attempt at a solution

∆T = (27E3)/(4186)(m)

Problem 2
1. The problem statement, all variables and given/known data

A thermos contains 159 cm3 of coffee at 91.1 °C. To cool the coffee, you drop two 12.3-g ice cubes into the thermos. The ice cubes are initially at 0 °C and melt completely. What is the final temperature of the coffee in degrees Celsius? Treat the coffee as if it were water.

V(coffee) = 159 cm^3
T(initial coffee) = 91.1°C
m (ice cubes) = 12.4 g @ 0°C
T(final coffee) = ?

2. Relevant equations
Q = mcΔT
Q = mLf

3. The attempt at a solution
I set the two equations equal to one another and used the Lf of 334 from Wikipedia for water. My answer was wrong though.

For problem 3:
1. The problem statement, all variables and given/known data
A 0.280-kg piece of aluminum that has a temperature of -154 °C is added to 1.2 kg of water that has a temperature of 2.1 °C. At equilibrium the temperature is 0 °C. Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.

m (Al) = .280 kg
T (Al) = -154°C
m (H2O) = 1.2 kg @ 2.1 °C

2. Relevant equations
Q = mcΔT
Q = mLf

3. The attempt at a solution
I set them equal to one another and tried to solve this similar to the previous one, but i'm really stumped.

Thank you, any help is appreciated.

2. Jan 27, 2012

### PeterO

3. Jan 27, 2012

### platachog

Well, for the 4186, I don't quite remember how I got that as the specific heat of water. I think I may have seen it in an example problem in the book. Doesn't the S.H. of H2O = 1?

I don't know how wrong my answer was, I just know it was more than 2% wrong. I may have forgotten about the water that is 0°C, but don't know how to account for it.

4. Jan 27, 2012

### platachog

The problem didn't provide me with the heat capacities, but even if I had them, I don't know how to use them.

5. Jan 27, 2012

### SammyS

Staff Emeritus
Do you know the units for heat capacity ?

Like most measured quantities in physics, without units, the number doesn't mean anything.

Yes, the specific heat of water is 1, it's also 4186, also 4.186, etc.

6. Jan 27, 2012

### PeterO

You quoted the formula Q = mcΔT

A good start would be to identify what each of those symbols stand for - including units.

Even temperature can be measured on two scales, so you have to make sure you are using matching units for each quantity.