Question

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.

Answer 1

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

Step-by-step 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.

`Heat_{\text{absorbed}}=Heat_{\text{released}}`

The equation used to calculate heat released or absorbed follows:

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]` ......(1)

where,

q = heat absorbed or released

`m_1` = mass of metal = 24.87 g

`m_2` = mass of water = 76.12 g

`T_(final)` = final temperature = 28.2°C

`T_1` = initial temperature of metal = 104.0°C

`T_2` = initial temperature of water = 25.2°C

`c_1` = specific heat of metal = ?

`c_2` = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

`24.87* c_1* (28.2-104)=-[76.12* 4.186* (28.2-25.2)]`

`c_1=0.507J/g^oC`

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