Specific Heat Capacity of a cylinder in boiling water

[DavisonH]

Homework Statement

Finding the specific heat capacity of an unknown metal.

A cylinder was placed into boiling water, originally at 24.2 C. The heat transferred from the metal, and the water then was 29.5 C.

Original Temperature-24.2 C
Temperature after the metal has been placed into the water.-29.5 C

m=mass of water (100 g)
C=Specific Heat of Water (1 C)
deltaT=Change of Temperature in water=5.3 (29.5-24.2)

Metal

M=Mass of metal (86.79 g)
deltaT=? (I think it is 5.3, equivalent to the change of temperature in the water)
C=?

I am not entirely sure what I am doing wrong. I thought that the heat gained by the water is equivalent to the heat lost by the metal, therefore the change

Homework Equations

Q=mCdeltaT

Q/m*deltaT=c

The Attempt at a Solution

First I found the heat gained by the water by using the formula Q=mCdeltaT.

Where
m=mass of water (100 g)
C=Specific Heat of Water (1 C)
deltaT=Change of Temperature in water=5.3 (29.5-24.2)

Q=100x1x5.3

Therefore Q=530

Next, I used the Q to calculate the specific heat capacity of the unknown metal.

Q=mCdeltaT, with the variables changed to Q/m*deltaT=c
Where
Q=Heat Gained by Water/Lost by metal (530)
m=Mass of metal (86.79 g)
deltaT=This is what I am unsure about. Would the heat lost by the metal be equivalent to the heat gained by the water? Which would equal to 5.3 C.

If I calculate using that assumption for Delta T the answer equals 1.15 c/ g cal which is quite high for the specific heat capacity of a metal.

Was my process correct?

 

[Redbelly98]

Welcome to Physics Forums

Think about this: what is the initial temperature of the boiling water?

 

[DavisonH]

Thanks!

The initial temperature of the boiling water is 98 C.

Then I think it would go like this

(86.79) C (100 C - 29.5C) = (100g) (1 cal/g-C)(5.3C)

(86.79) (70.5C) C = 530

6118.7 C = 530

C = 0.0866 cal/g-C

 

[Redbelly98]

Getting there, but there are still some problems.

Thanks!

The initial temperature of the boiling water is 98 C.

Well, you're close, but ... hey, come on, what temperature does water boil at?

Then I think it would go like this

(86.79) C (100 C - 29.5C) = (100g) (1 cal/g-C)(5.3C)

Hmmm. Okay, one side of that equation refers to the water, and the other side of the equation refers to the metal. Which is which?

 

[DavisonH]

The boiling point of water is 100 C, but the water measured in the lab as at 98 C.

The left side is the metal, the right the water.

 

[Redbelly98]

The boiling point of water is 100 C, but the water measured in the lab as at 98 C.

Understood, thanks for clarifying.

(86.79) C (100 C - 29.5C) = (100g) (1 cal/g-C)(5.3C)
.
.
.
The left side is the metal, the right the water.

The metal does not go from 100C to 29.5C. The temperature change of the water is not 5.3C, that is ΔT for the metal.
Fix those temperatures, then you should have it ... and remember, the water starts out at 98C.