Question

A calorimeter has aluminum inner cup of mass 120 gram containing 200 ml water at temperature 20 degree Celsius. Brass piece with mass 100 gram is heated to 100 degree Celsius, and then immersed in the calorimeter. Calculate the final temperature of the system. The specific heat of brass is 0.09 cal/(gramXdeg.C) . The additional necessary data are provided in the text.
2) A chunk of 22 gram ice is pulled from a freezer keeping temperature -25 deg. Celsius, and placed in an identical calorimeter as the one described in problem 1. The final temperature of the system is measured to be 10 deg. Celsius. What is the experimental latent heat of fusion of ice? The additional necessary data are provided in the text.

Answer 1

Answer:

1. Final temperature of the system = 23.06 oC

2. Experimental latent heat of fusion of ice = 80.41 Cal/g

Explanation:

1. Heat energy is usually transferred from a hotter to a colder body.

Heat lost by Brass = Heat gained by aluminum calorimeter + heat gained by liquid water in the calorimeter.

But,

Heat, H = mc∆T

Where m = mass of the substance, c = specific heat capacity of the substance, ∆T = change in temperature.

For Brass, H = 100 x 0.09( 100 – final temperature )

H = 9 ( 100 – final temperature )

H = 900 – 9final temp.

For aluminum calorimeter,

H = 120 x 0.22( final temperature – 20 )

H = 26.4( final temp. – 20 )

H = 26.4final temp. – 528

For liquid water in the calorimeter,

200ml = 200g, so,

H = 200 x 1( final temperature – 20 )

H = 200finaltemp. – 4000

Equating heat loss with heat gain,

900 – 9final temp. = 26.4final temp. – 528 + 200final temp. – 4000

200final temp. + 26.4final temp. + 9final temp. = 900 + 528 + 4000

235.4final temp. = 5428

Final temp. = 5428 / 235.4

Final temp. = 23.06 oC.

2. Since the total energy of the system is conserved,

Heat lost by aluminum calorimeter + Heat lost by water in the calorimeter + Heat required to raise the temperature of ice from -25 oC to 0 oC + Heat required to melt ice at 0 oC + Heat required to raise the temperature of liquid water from 0 oC to 10 oC = Zero

120 x 0.22(10-20) + 200 x 1(10-20) + 22 x 0.5(0-(-25)) + 22 x L + 22 x 1(10-0) = 0

- 264 – 2000 + 275 + 22L + 220 = 0

22L = 264 + 2000 – 275 – 220

22L = 2264 – 495

22L = 1769

L = 1769 / 22

L = 80.41 Cal/g

Where L = latent heat of fusion of ice.

Answer 2

Answer: Final Temperature= 40.57 deg C; Latent Heat of Fusion of Ice= 80.14 cal/g

Explanation:

The system is said to be isolated to the surroundings. All energy must remain inside the system. We can state that no loss of energy takes place, thus,