Question

A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperature of the system is 25.0 °C. Determine the specific heat of the metal.

Answer 1

Answer: The specific heat of metal is 0.821 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 = 30 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = 25°C

$T_1$ = initial temperature of metal = 110°C

$T_2$ = initial temperature of water = 20.0°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:

$30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]$

$c_1=0.821J/g^oC$

Hence, the specific heat of metal is 0.821 J/g°C