# Heat question

1. May 15, 2015

### goonking

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

2. Relevant equations
Q = Lm (L is latent heat of fusion)
Q = c m ΔT (c is specific heat) (m is mass)

3. The attempt at a solution

equating the 2 equations : Lm = c m ΔT

the m's cancel out

L = c ΔT

feels like something went wrong at this point

2. May 15, 2015

### Orodruin

Staff Emeritus
You will need to take into account that the ice will continue to cool the coffee after melting. What will be the relation which tells you how much heat is released when the ice melts and gains 26 °C?

Important side note: I do not know if you did the image of the problem yourself, but C is a unit of charge, not of temperature. The unit of temperature is °C.

3. May 15, 2015

### goonking

Q = L m

so for example
(1 kg of ice ) ( 3.3x105 J/kg) = 3.3 x 105 Joules

also, are you referring to the c that denotes specific heat?

4. May 15, 2015

### Orodruin

Staff Emeritus
No, this is the heat required when melting the ice only. The ice will also need to be heated to 26 °C.

No, I am referring to what is stated in the problem. For example 26 C is a (quite large) charge, not a temperature - 26 °C is a temperature.

5. May 15, 2015

### goonking

ah, it was probably forgotten , sorry about that.

Q = c m Δ T

where Δ T is (26 - 0) = -26 °C

6. May 15, 2015

### Orodruin

Staff Emeritus
So what is the total heat required to melt the ice and heat it to 26 °C?

7. May 15, 2015

### goonking

ok, so we have 200 mL of coffee = 0.2 L

density of coffee is 1000 g/L

1000 = m / 0.2

m = 200 g of coffee = .2 kg

heat required to change coffee temperature from 80C to 26 C is:

Q = c m ΔT =( 4190 ) ( .2 kg ) ( 26 C-80 C)

= -45252 joules

so amount of ice i need is

Q = L m
45252 = 3.3x105 m

m = 0.137 kg of ice

correct?

Last edited: May 15, 2015
8. May 15, 2015

### Orodruin

Staff Emeritus
No, the ice is in the coffee so you will also need to heat it from 0 °C to 26 °C in order for the coffee (and ice, which is now water) to have a temperature of 26 °C. This requires more heat than simply melting the ice.

9. May 15, 2015

### goonking

don't we need the specific heat of ice to do this?

10. May 15, 2015

### Orodruin

Staff Emeritus
No, the ice is already at the melting point. After melting, it is water.

11. May 15, 2015

### goonking

isn't the ice getting heat from the coffee? we just needed ice to absorb heat from the coffee so the temperature of the coffee decreases from 80 C to 26 C?

12. May 15, 2015

### Orodruin

Staff Emeritus
If you put the amount of ice you quoted, you will end up with coffee at 26 °C mixed with water at 0 °C. The resulting mixture will obtain a temperature in between.

13. May 15, 2015

### goonking

oooh, so we actually didn't need that much ice. k, let me do some thinking

14. May 15, 2015

### goonking

Q = (mass ice )( cwater) ( [PLAIN]http://antoine.frostburg.edu/chem/senese/images/Delta.gifTmelted [Broken] ice)

so Q = massice (4190) ( 0C - 26C)

but we still have 2 unknowns, Q and mass of ice

Last edited by a moderator: May 7, 2017
15. May 15, 2015

### Orodruin

Staff Emeritus
This is now the heat required to heat up the water (previously ice) to 26 °C. So what is the heat required to melt the ice and then heat the resulting water to 26 °C? This must be equal to the heat released when cooling the coffee down to 26 °C.

Last edited by a moderator: May 7, 2017
16. May 15, 2015

### goonking

so we found out the water needs to release 45252 joules of heat to get it from 80 C to 26 C.

so the heat required to melt the ice and then heat the resulting water to 26 C = 45252 joules?

ok, so i googled the specific heat of ice , and it is 2093

placing it into this : Q = m c ΔT = 45252 joules = mice 2093 (26 C)

gives me a mass of 0.831 kg of ice

which sounds like too much ice to be realistic.

Last edited: May 15, 2015
17. May 15, 2015

### Orodruin

Staff Emeritus
As I said, you do not need the specific heat of ice, it is already at melting temperature and after melting it will be water.
You are also again neglecting the heat required to actually melt the ice. You must consider both the melting heat and the heat required to heat the resulting water at the same time.

18. May 15, 2015

### goonking

wait, if something is already at its melting temperature, we ignore the c in Q =m c Δ T?

19. May 15, 2015

### Orodruin

Staff Emeritus
No, but you do not need to change the temperature to melt it. You need to supply the melting heat. Then you need to raise the temperature of the water by 26 degrees.

I am sorry, but I cannot say it in more ways: You need to take two contributions into account, the heat required to melt the ice and the heat required to heat up the resulting water.

20. May 15, 2015

### goonking

ok , bear with me , so we know the water releases 45252 joules

so the heat required to melt the ice + the heal required to heat up the melted ice to 26 C = 45252

and heat required to melt the ice is Q1 = (mice) (3.3 x 105)

and heat required to heat up the melted ice to 26 C is Q = 4190 mice 26C

[(mice) (3.3x105 )] + [ (4190)(mice) (26)] = 45252

factoring out the m in for left side
m [(3.3x105 ) + (4190)(26C)] = 45252

m = .103 kg
correct?

Last edited: May 15, 2015