An insulated beaker contains 250g of water at 25 degrees celsius. Exactly 41.6g of a metal at 100 degrees celsius was dropped in the beaker. The final temperature of the water was 26.4 degrees celsius. Assuming that no heat is lost in any other way, calculate the specific heat of the metal.

Question

I know the equation for specific heat is q= m X h X (Tf-Ti). I am just not sure where to plug everything in at.

I know that the final temp is 26.4 and the inital is 25.

Answer 1

This is how I solved it.

M (water) = 250g
T (initial water) = 25 + 273 = 298K
M (metal) = 41.6g
T (initial metal) = 100 + 273 = 373K
T (final water & metal) = 26.4 + 273 = 299.4
C (water) = 4.184 (specific heat of water)
C (metal) = x

[4.184 * 250g (299.4 - 298)] + [x * 41.6g (299.4-373)]

I get 0,478 J g . K for x (specific heat of metal)

Answer 2

Ok thank you for your help!!