A 24.87 gram sample of a metal at 104.0ºC was added to 76.12 grams of water at 25.2ºC in a perfectly insulated calorimeter. The final temperature of the metal-water mixture was 28.2ºC. Calculate the specific heat capacity of the metal using the data. Type your work and answer below. Make sure to include a unit on the final answer.

The specific heat of metal is 0.507 J/g°C

Explanation:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

The equation used to calculate heat released or absorbed follows:

      ......(1)

where,

q = heat absorbed or released

= mass of metal = 24.87 g

= mass of water = 76.12 g

= final temperature = 28.2°C

= initial temperature of metal = 104.0°C

= initial temperature of water = 25.2°C

= specific heat of metal = ?

= specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

Hence, the specific heat of metal is 0.507 J/g°C

wher is the rest of the question