Question

A 55.0 g piece of metal has an initial temperature of 75.2 °C. The piece of metal is immersed in a beaker containing 106.9 g of water at 23.5°C. When the two substances reach thermal equilibrium, the final temperature is 28.8°C. What is the specific heat of the metal?

Answer 1

c(H2O)=4200 J/kg*°C

c=Q/m‎ΔT;

Q=c*m*‎ΔT=4200*0.1069*(28.8-23.5)=2379.594 J

c(Me)=2379.594/(0.055*(75.2-28.8)=932.44 J/kg*°C

Answer: 932.44 J/kg*°C