Question

A 0.0625 kg ingot of metal is heated to 213◦C
and then is dropped into a beaker containing
0.4 kg of water initially at 20◦C.
If the final equilibrium state of the mixed
system is 22.4◦C, find the specific heat of
the metal. The specific heat of water is
4186 J/kg ·◦ C.
Answer in units of J/kg ·◦ C.

Answer 1

$$mc(T_m-T)=Mc_w(T-T_w)$$

$$(0.0625)(213-22.4)c=(0.4)(4186)(22.4-20)$$

$$c=337\frac{J}{kg \degree C}$$