Posted by **bon** on Tuesday, January 23, 2007 at 1:55pm.

When 80.0 grams of a certain metal at 90.0 °C was mixed with 100.0 grams of water at 30.0 °C, the final equilibrium temperature of the mixture was 36.0 °C. What is the specific heat (J/g°C) of the metal?

I suppose I would use this formula:

q= mcÄt

m= mass(g)

q= enthalpy change (j)

c= specifice heat capacity(J/g°C)

Ät= temperature change(°C)

not quite sure how to approach this question!?!?!?!

The sum of the heats added is zero (one is lost, so it is negative).

Heataddedwater + heataddedmetal=0

mw*cw*(Tf-Tiw) + mm*Cm*(Tf-Tim)=0

solve for Cm, the specific heat of the metal.

- chem+ help! -
**jacqueline **, Saturday, February 6, 2016 at 2:09pm
0.581J/g c

## Answer This Question

## Related Questions

- Chemistry - 100 grams of glucose is mixed with 500 grams of water. 10 grams of ...
- Physics - 400 grams of a metal at 300 C is added to 100 grams of of water at 20...
- physics - A 200 gram copper calorimeter contains 400 grams of water at 25 C. A ...
- Physics - One hundred twenty grams of metal at 88 degree Celsius is poured into ...
- chemistry - if 50 grams of a substance with temperature of 20 degrees C and a ...
- Physics - 120 grams of metal at 88 degree celcius is poured into a 70 grams ...
- chemistry - a mixture of 100 grams of water is mixed with 100 grams of sodium ...
- Physics - If you mix 100 grams of water at 20 degree centigrade with 100 grams ...
- Chemistry - A 30.0 gram sample of water at 280. K is mixed with 50.0 grams of ...
- chemistry - if 10 grams of metal A changes the temperature of 10 grams of water...

More Related Questions