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.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

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

$47.1g\times c_1\times (24.5-99)^oC=-120g\times 4.18J/g^oC\times (24.5-21.4)^oC$

$c_1=0.44J/g^oC$

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 $0.44J/g^oC$).