Question

A 47.1 g sample of a metal is heated to 99.0°C and then placed in a calorimeter containing 120.0 g of water (c = 4.18 J/g°C) at 21.4°C. The final temperature of the water is 24.5°C. Which metal was used?

Answer 1

Answer : The metal used was iron (the specific heat capacity is
`0.44J/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 metal = ?

`C_1` = specific heat of water =
`4.18J/g^oC`

`m_1` = mass of metal = 47.1 g

`m_2` = mass of water = 120 g

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

`T_1` = initial temperature of metal =
`99^oC`

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

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

`47.1g* c_1* (24.5-99)^oC=-120g* 4.18J/g^oC* (24.5-21.4)^oC`

`c_1=0.44J/g^oC`

Form the value of specific heat of metal, we conclude that the metal used in this was iron.

Therefore, the metal used was iron (the specific heat capacity is
`0.44J/g^oC`).