Question

A silver block, initially at 56.1 ∘C, is submerged into 100.0 g of water at 24.0 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 28.0∘C. What is the mass of the silver block?

Answer 1

Answer: The mass of the silver block is 256 grams

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 silver = ?

$m_2$ = mass of water = 100.0 g

$T_{final}$ = final temperature = $28.0^0C$

$T_1$ = temperature of silver = $56.1^oC$

$T_2$ = temperature of water = $24.0^oC$

$c_1$ = specific heat of silver = $0.233J/g^0C$

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

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

$-m_1\times 0.233\times (28.0-56.1)=[100.0\times 4.184\times (28.0-24.0)]$

$m_1=256g$

Therefore, the mass of the silver block is 256 grams