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Physics Specific Heat Problems

physics specific heat problems
Solving specific heat—physics problem. HELP!?

Please help me solve this problem. I’ve been working on this for like 3 hours and have not yet come up with an answer. If you get an answer, can you please show your work?
—
A hot, just-minted copper coin is placed in
116 g of water to cool. The water temperature
changes by 8.00 °C and the temperature of the
coin changes by 87.3 °C.

What is the mass of the coin? Disregard
any energy transfer to the water’s surround-
ings and assume the specific heat of copper
is 387 J/kg°C. The specific heat of water is
4186 J/kg°C.

Answer in units of g.
—

Thanks so much.

OK … here we go:

The energy absorbed by the water is:

E = 4184J/(kg*C)*8C*0.160kg = 495.36J

For the copper, we want to know the mass so let’s multiply its specific heat by what we know and see what happens. Note: we must use the inverse of the specific heat to get the units right and get an answer in grams. Lets try:

(kg*C/387J)*(495.36J/87.3C) = 0.014662kg or 14.662g

Lets check it just to make sure.

E = 387J/(kg*C)*87.3C*0.014662kg = 495.36J

Both the copper and the water exchanged the same number of joules!

The answer is 14.662 grams of copper.

Best,
Dave

Physics: Thermodynamic processes (1)

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Specific Heat Problems Physics

specific heat problems physics
physics specific heat question?

just a little question about physics.
in a physics specific heat problem, why don’t you need to convert Kelvins into Celsius degrees?
For example, Find the amount of heat required to raise the temperature of 460 g of silver from 273 K to 300 K.
The equation is Q=mct with t being the change in temperature. The heat required is 234 J/kgC.
The book says that the change in temperature is 27 K and that’s the same as 27 Celsius degrees. I don’t understand this. Shouldn’t you change the 27 K into Celsius degrees by subtracting it from 273?

Although the Celsius and Kelvins scale differ in “where 0 is”,a degree in each scale has the same “size” . So when you only need differences in temperature, it does not matter numerically if you use Kelvin or Celsius.
In formula: K1 = C1 + 273, K2 = C2 + 273 , so
K2 – K1 = C2 + 273 – ( C1 + 273) = C2 – C1

Physics #2 – Density & Specific Heat.wmv