A 13.8 grams piece of zinc was heated to 99.8°C in boiling water and then dropped into a beaker containing 45.0 g of water at 25.0 °C.  When the water and metal come to thermal equilibrium, the temperature is 27.1°C.  What is the specific heat capacity of zinc?

- q lost (metal) = ms(tf-ti) =- 13.8*s*(27.1-99.
q gained (water) = ms(tf-ti) = 45*4.18*(27.1-25)
- q lost (metal)= q gained (water)

Solve it for x