Finding Specific Heat of oil with an electric heating coil

[Moose100]

Homework Statement

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?

Homework Equations

Q=mc(deltaT)
H=Q/change in time

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!

 

[tiny-tim]

Welcome to PF!

Hi Moose100! Welcome to PF!

what is the specific heat of the coil?

(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)

 

[Moose100]

Yes the oil! I am 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.

 

[tiny-tim]

… I have looked at conversions and such and my answer comes up short still.

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

then we help you ​

 

[Moose100]

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

then we help you ​

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 I am still new.

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

Why am I .04 short?

 

[tiny-tim]

Hi Moose100!

(just got up :zzz:)

15120J-600J

no, it's not 600J, it's (I think) 600 cal

 

[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*CStill 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 I am not just walking in here for an answer.

 

[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.

 

[Moose100]

Using time to convert into equivalent weight? How so?

 

[Moose100]

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

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

 

[tiny-tim]

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

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

yes, the joules are correct…

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

now convert that to cals

 

[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(?))

 

[tiny-tim]

15120j/4.184j/cal=3613.767 cal

ok, stop there…

now what is the definition of a cal?

 

[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.

 

[tiny-tim]

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

oh come on …

how much water?

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

 

[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.

 

[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

 

[Moose100]

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

 

[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

 

[tiny-tim]

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

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?​

 

[Moose100]

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?​

Q of water=20g(30C)(1cal/g)=600cal
Q of oil= 380g(c)(30C)=11.400c

where c is the specific heat of oil.

we already know the heat transfer of the coil 3613.77cal

that equals the sum of the oil and water.

solve for c.

I made a mistake when I wrote the problem we are finding specific heat of the oil.

 

[tiny-tim]

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

c=0.31cal/g*C

I don't know how you got 0.31 from that equation, I get 0.26

(but a more logical way would be to say that the water equivalent is 3613/30 = 120.5,

subtract the 20, = 100.5,

so the 380g of oil has a water equivalent of 100.5, so the specific heat is …)

 

[Moose100]

I don't know how you got 0.31 from that equation, I get 0.26

(but a more logical way would be to say that the water equivalent is 3613/30 = 120.5,

subtract the 20, = 100.5,

so the 380g of oil has a water equivalent of 100.5, so the specific heat is …)

It's a rounding error. The bold is what I corrected in my calulations(i.e. not rounding at all) to FINALLY get 0.26cal

Before when I rounded I got 0.31

The thing is I don't necessarily know what heat equivalent is.

What is the formula for what you are doing there? Whats the units j/C??

 

[tiny-tim]

It's a rounding error. The bold is what I corrected in my calulations(i.e. not rounding at all) to FINALLY get 0.26cal

Before when I rounded I got 0.31

There's no way that rounding can make the difference between .26 and .31 …

you must have made a mistake on your calculator …

try it again

 

[Moose100]

There's no way that rounding can make the difference between .26 and .31 …

you must have made a mistake on your calculator …

try it again

Lol it is actually.

Try it!