Question

4000J of heat is applied to 1.5 silver pendant initially at temperature of 150°C. Determine it's final temperature (latent heat=336J/Kg, heat capacity = 233J/Kg.K)

Answer 1

From the conservation of energy:

$$Q=mc(T-T_0)$$

$$4000=1.5(233(T-15))$$

$$T = 26.4^\circ C$$