Latent Heat, Some Density, and God knows what else

In summary: From that equation, you should be able to work out the units of cYou also need to figure out how much ice you have, and the latent heat of fusion for ice.Once you have that, you can set up an equation for the amount of heat lost by the coffee (and any ice that may have melted) and the amount of heat gained by the ice cubes (in melting them and warming them up to the final temperature) and then equate these two quantities.This should give you an equation that you can solve for the final temperature of the coffee.In summary, the conversation is discussing problems in physics, specifically related to heating and cooling systems and the properties of water. The first problem involves calculating the maximum temperature of
  • #1
platachog
3
0
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:

Homework Statement


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) = ?

Homework Equations


Q = mcΔT
ρ = m/v

The Attempt at a Solution



∆T = (27E3)/(4186)(m)


Problem 2

Homework Statement



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) = ?

Homework Equations


Q = mcΔT
Q = mLf

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:

Homework Statement


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

Homework Equations


Q = mcΔT
Q = mLf

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.
 
Physics news on Phys.org
  • #2
platachog said:
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:

Homework Statement


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) = ?

Homework Equations


Q = mcΔT
ρ = m/v

The Attempt at a Solution



∆T = (27E3)/(4186)(m)

Where did you get your figure of 4186 from? Not saying it is wrong, just want to see how you got it.
Did you allow for 4 showere heads each with the given flow rate?
When you get a final ΔT don't forget that that is how much hotter than 10°C the water is


Problem 2

Homework Statement



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) = ?

Homework Equations


Q = mcΔT
Q = mLf

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.

How wrong was your answer? If you were only out by a few degrees you may have accounted for the ice melting, but forgot about the melted ice [water at 0o then warming to the final temperature


For problem 3:

Homework Statement


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

Homework Equations


Q = mcΔT
Q = mLf

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.

I don't see any reference to the specific heat capacity of either Al or water, not the latent heat of fusion for ice?

Thank you, any help is appreciated.

Comments above.
 
  • #3
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
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
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
platachog said:
The problem didn't provide me with the heat capacities, but even if I had them, I don't know how to use them.

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.
 

What is latent heat?

Latent heat is the energy required to change the phase of a substance without changing its temperature. This means that when a substance changes from a solid to a liquid or from a liquid to a gas, energy is either absorbed or released in the form of latent heat.

How is latent heat measured?

Latent heat is measured in units of joules (J) per kilogram (kg) or calories (cal) per gram (g). The specific latent heat is the amount of energy required to change the phase of 1kg of a substance at a constant temperature.

What is density?

Density is a measure of how much mass is contained in a given volume. It is calculated by dividing the mass of an object by its volume. The unit of density is typically grams per cubic centimeter (g/cm3) for solids and liquids, and grams per liter (g/L) for gases.

How does density affect the behavior of substances?

Density plays a significant role in the behavior of substances. Less dense substances will float on top of denser substances, and substances with similar densities will mix together. Density also affects the boiling and melting points of substances, as well as their ability to dissolve in other substances.

What is the relationship between density and temperature?

Generally, the density of a substance decreases as its temperature increases. This is because as temperature increases, the molecules of a substance have more energy and move faster, causing them to take up more space and decrease in density. However, this relationship may not hold true for all substances, as some substances may have a different behavior due to their atomic or molecular structure.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
983
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top