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

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 = 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* c_1* (25-110)=-[100* 4.186* (25-20)]`

`c_1=0.821J/g^oC`

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