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

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:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$       ......(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:

$24.87\times c_1\times (28.2-104)=-[76.12\times 4.186\times (28.2-25.2)]$

$c_1=0.507J/g^oC$

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