# Very Basic Physics that I'm struggling on

Very Basic Physics....that I'm struggling on...

## Homework Statement

In an experiment to find the temperature of a Bunsen flame, a 0.2kg brass mass is heated in the flame. The hot mass is then placed in a beaker containing 0.2 kg of water. The temperature of the water rises from 20C to 80C.
a)what is the rise in temperature of the water?
b)what is the heat energy gained by the water?
c)assuming the temperature of the mass to be T, what is the fall in temperature of the mass?
d)what is the heat lost by the mass?
e)assuming that the heat lost by the mass is the same as the heat gained by the water, find the temperature, T.

## Homework Equations

H=m*c*rise in temperature

## The Attempt at a Solution

I think I have the first four. I did...
a) 80C-20C = 60C
b) H=0.2*4200*60 = 50400J
c) H/mc=change in temp. therefore, -50400/0.2*370=change in temp. therefore, the change in temp. is -681C
d) I just assumed that the the heat lost by the mass is the same as the heat gained by the water, which is 50400J.
e) This one has me completely stuck...

Last edited:

Related Introductory Physics Homework Help News on Phys.org

In an experiment to find the temperature of a Bunsen flame, a 0.2kg brass mass is heated in the flame. The hot mass is then placed in a beaker containing 0.2 kg of water. The temperature of the water rises from 20C to 80C.
a)what is the rise in temperature of the water?
b)what is the heat energy gained by the water?
c)assuming the temperature of the mass to be T, what is the fall in temperature of the mass?
d)what is the heat lost by the mass?
e)assuming that the heat lost by the mass is the same as the heat gained by the water, find the temperature, T.

## The Attempt at a Solution

I think I have the first four. I did...
a) 80C-20C = 60C
b) H=0.2*4200*60 = 50400J
c) H/mc=change in temp. therefore, -50400/0.2*370=change in temp. therefore, the change in temp. is -681C
d) I just assumed that the the heat lost by the mass is the same as the heat gained by the water, which is 50400J.
e) This one has me completely stuck...
you already computed the temperature loss of the brass. The brass ends up at the same temperature of the water after losing 681 degrees