heat lost by Mo + heat gained by H2O = 0
[mass Mo x specific heat Mo x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)]= 0
specific heat Mo is the only unknown.
A 237 g piece of molybdenum, initially at 100.0°C, is dropped into 244 g of water at 10.0°C. When the system comes to thermal equilibrium, the temperature is 15.3°C. What is the specific heat capacity of molybdenum?
3 answers
0.2695J
e213