In an experiment, 21.5 g of metal was heated to 98.0°c and then quickly transferred to 150.0 g of water in a calorimeter. the initial temperature of the water was 21.5°c, and the final temperature after the addition of the metal was 32.5°c. assume the calorimeter behaves ideally and does not absorb or release heat.
what is the value of the specific heat capacity (in j/g•°c) of the metal? j/g•°c

The value of the specific heat capacity is 4.89 J/g°C

Explanation:

First of all, let's make the equation.

Assume the calorimeter behaves ideally and does not absorb or release heat, this phrase explains that all the heat released the metal, was gained by the water so:

Qmetal = metal mass . C metal . ΔT

Q water = water mass . C H2O . ΔT

Q metal = Q water

21.5 g . C metal (32.5 °C - 98°C ) = 150 g . 4.18 J/g°C (32.5°C - 21.5°C)

21.5 g . C metal . -65.5 °C = 150g . 4.18 J/g°C . 11°C

21.5 g . C metal . -65.5 °C = 6897 J

C metal =  6897 J /21.5 g . 65.5°C

C metal = 4.89 J/g°C

Notice that specifc heat is a constant. It can be negative. NEVER

We use the absolute value. |x|