Question

What is the final temperature of a system if 13.00 g of gold at 90.0°C is placed in 13.00 g of water at 26.00°C? The molar heat capacity of gold is 25.41 J/(mol · °C) and the heat capacity of water is 4.18 J/(g · °C).

Answer 1

Answer : The final temperature of the system is, $27.9^oC$

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 gold = $0.129J/g.^oC$

Note : Molar heat capacity of gold = $25.41J/mol.^oC$

Heat capacity of gold = $\frac{25.41J/mol.^oC}{197g/mol}=0.129J/g.^oC$

$c_2$ = heat capacity of water = $4.18J/g.^oC$

$m_1$ = mass of gold = 13.00 g

$m_2$ = mass of water = 13.00 g

$T_f$ = final temperature of system = ?

$T_1$ = initial temperature of gold = $90.0^oC$

$T_2$ = initial temperature of water = $26.00^oC$

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

$13.00g\times 0.129J/g.^oC\times (T_f-90.0)^oC=-13.00g\times 4.18J/g.^oC\times (T_f-26.00)^oC$

$T_f=27.9^oC$

Therefore, the final temperature of the system is, $27.9^oC$