Question

A 29.950 g sample of an unknown metal is heated to 100.0 °c. The metal is then poured into 50.0 g of water in an insulated coffee cup calorimeter. The temperature of water rises from 25.0 ºc to 27.7 ºc. What is the specific heat of the metal?

Answer 1

General metal is highly heat temperature

Answer 2

Answer : The specific heat of metal is,
`0.261J/g^oC`

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`c_1` = specific heat of unknown metal = ?

`c_2` = specific heat of water =
`4.18J/g^oC`

`m_1` = mass of unknown metal = 29.950 g

`m_2` = mass of water = 50.0 g

`T_f` = final temperature of water =
`27.7^oC`

`T_1` = initial temperature of unknown metal =
`100^oC`

`T_2` = initial temperature of water =
`25.0^oC`

Now put all the given values in the above formula, we get

`29.950g* c_1* (27.7-100)^oC=-50.0g* 4.18J/g^oC* (27.7-25.0)^oC`

`c_1=0.261J/g^oC`

Therefore, the specific heat of metal is,
`0.261J/g^oC`