Question

A 125-g piece of metal is heated to 288 °C and dropped into 85.0 g of water at 12.0 °C. The metal and water come to the same temperature of 24.0 °C. What is the specific heat, in J/g °C, of the metal?

Answer 1

heat lost by metal is gained by water

125 g * x J/g⋅°C * (288 - 24.0) ºC = 85.0 g * 4.18 J/g⋅°C * (24.0 - 12.0) ºC

x is the specific heat of the metal

Answer 2

So I solve for x?

Then, the SH is .129 J/g °C?

Thank you for helping me out!