Question

A 31.5 g wafer of pure gold initially at 69.4 ∘C is submerged into 63.4 g of water at 27.4 ∘C in an insulated container.

Answer 1

Answer: The final temperature of both substances at thermal equilibrium is 301.0 K

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,  

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$         .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of gold = 31.5 g

$m_2$ = mass of water = 63.4 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of gold = $69.4^oC=342.4K$

$T_2$ = temperature of water = $27.4^oC=300.4K$

$c_1$ = specific heat of gold = $0.129J/g^0C$

$c_2$ = specific heat of water= $4.184J/g^0C$

Now put all the given values in equation (1), we get

$-31.5\times 0.129\times (T_{final}-342.4)=[63.4\times 4.184\times (T_{final}-300.4)]$

$T_{final}=301.0K$

The final temperature of both substances at thermal equilibrium is 301.0 K