# Homework Help: Finding Specific Heat of oil with an electric heating coil

1. Jan 22, 2012

### Moose100

1. The problem statement, all variables and given/known data
To determine the Specific heat of an oil , an electrical heating coil is placed in a calorimeter with 380g of the oil at 10C. The coil consumes energy(and gives off heat) at the rate of 84W. After 3min the oil temperature is 40C. If the water equivalent of the calorimeter and coil is 20g, what is the specific heat of the coil?

2. Relevant equations
Q=mc(deltaT)
H=Q/change in time

3. The attempt at a solution

Q=Ht=3min(60s)(84W)=15120J
15120=380cm(c)(40-10C)+20cm(1)(40-10C)

I have been at this problem for HOURS the answer is 0.26 cal/g*C.
Also I don't really know what "water equivalent of the calorimeter and coil" means.

I would think the final temp after 3min of heating is the total heat but obviously I am wrong PLEASE HELP!

2. Jan 22, 2012

### tiny-tim

Welcome to PF!

Hi Moose100! Welcome to PF!
(you mean the oil?)

cal is a weird non-SI unit

you need to look up its definition and find the conversion ratio for joules or watts

(and i think "water equivalent" means that it heats up at the same rate as 20g of water would)

3. Jan 22, 2012

### Moose100

Yes the oil! Im so sorry lol!

I have tried to do this problem every which way I could. Please help. Lol I have looked at conversions and such and my answer comes up short still.

4. Jan 22, 2012

### tiny-tim

This forum's rules mean you have to show us what you've done (including that conversion factor) …

5. Jan 22, 2012

### Moose100

I have? I think:

Q=Ht=3min(60s)(84W)=15120J
15120=380cm(c)(40-10C)+20cm(1)(40-10C)

Do you mean show you the conversions? Sorry Im still new.

15120J-600J=14520J/30C(380)(4.184J/Cal)=.30cal

Why am I .04 short?

6. Jan 23, 2012

### tiny-tim

Hi Moose100!

(just got up :zzz:)
no, it's not 600J, it's (I think) 600 cal

7. Jan 23, 2012

### Moose100

Q=Ht=3min(60s)(84W)=15120J
15120J=0.380kg(c)(40-10C)+0.020kg(4.184J/kg*C)(40-10C)
(15120J-2.51J )/11.4kg*C =15117J/kg*C/4.184J=3613cal/kg*C

Still wrong. I just feel like I am doing something wrong overall I am still lost and very frustrated could you please give me something? I think I have proven that Im not just walking in here for an answer.

Last edited: Jan 23, 2012
8. Jan 23, 2012

### tiny-tim

Do it systematically …

start from one end and go towards the other end.​

Start with the number of J, convert that into cal, then use the time to convert that into the total equivalent weight of water.

9. Jan 23, 2012

### Moose100

Using time to convert into equivalent weight? How so?

10. Jan 23, 2012

### Moose100

I used 85W and the time 3min to convert that to Joules

85W(3min)(60s/min)=15120J

11. Jan 23, 2012

### tiny-tim

yes, the joules are correct…

(btw, no capital "J" in "joules" )

now convert that to cals

12. Jan 23, 2012

### Moose100

15120j/4.184j/cal=3613.767 cal

Qof water= 20g(1cal/g)(40C-10C)=600cal

Qof oil 380g(c of oil)(40-10C)=11400c(units are cal(?))

Last edited: Jan 23, 2012
13. Jan 23, 2012

### tiny-tim

ok, stop there…

now what is the definition of a cal?

14. Jan 23, 2012

### Moose100

The amount of energy required to rais the temperature of water 1C.

In one calorie there are 4.184j. In one Calorie there are 1000calories. So 1Calorie=1000kcal.

15. Jan 23, 2012

### tiny-tim

oh come on

how much water?

(and next, what's the definition of specific heat?)

16. Jan 23, 2012

### Moose100

Oh crap. I was thinking it but didn't type it my fault.

1g of water a calorie is the amount of energy required to raise one gram one degree.

Specific heat is the amount of energy required to raise a gram(or kg) of a specific substance 1 degree.

17. Jan 23, 2012

### tiny-tim

ok, now find the water equivalent of the whole thing (the oil plus the calorimeter and coil)

the water equivalent is how much water there would be if it was all made of water, and if the same heat produced the same temperature rise

18. Jan 23, 2012

### Moose100

So..Im sorry im still fuzzy on this...The water equivalent isn't what those conversions show?

19. Jan 23, 2012

### Moose100

3613.76673cal=600cal+11,400c
(3613.76673cal-600cal)/11400g*C = c-specific heat of oil

c=0.31cal/g*C (It IS rounding error!)

this is off from what the book has as an answer. I think this is what I was getting before is this simply rounding error?

If I hadn't of dropped some decimals I woulda got 0.26cal/g*C

Last edited: Jan 23, 2012
20. Jan 23, 2012

### tiny-tim

you've lost me

where does 600 or 11,400 come from?

the question says …
To determine the Specific heat of an oil , an electrical heating coil is placed in a calorimeter with 380g of the oil at 10C. The coil consumes energy(and gives off heat) at the rate of 84W. After 3min the oil temperature is 40C. If the water equivalent of the calorimeter and coil is 20g, what is the specific heat of the coil?​