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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.


Physics: Thermodynamic processes (1)

Specific Heat Problems Chemistry

specific heat problems chemistry
AP chemistry specific heat problem?

a coffee cup calorimeter contains 50 g of water at 60.51 degrees celcius. a 12.4g piece of graphite at 24.21 degrees celcius is placed in the calorimeter. the final temperature of the water and the carbon is 59.02 degrees celcius. calculate the specific heat of carbon. the specific heat of water is 4.18 j/g*C

thank you

59.02C – 24.21C = 34.81degC temperature change for graphite

60.51C – 59.02C = 1.49degC temperature change for water

(50gH2O)(1.49degC)(4.18J/gH2O) = 311.41 J heat lost by water and gained by graphite (G)

(311.41J)/(12.4gG)(34.81degC) = 0.7214 J/g-degC

Specific Heat Capacity Sample Problem 1