Question

Suppose you heat a metal object with a mass of 31.7 g to 96.5 oC and transfer it to a calorimeter containing 100.0 g of water at 17.3 oC. The water and metal reach a final temperature of 24.6 oC. What is the specific heat of the metal in J/g-oC?

Answer 1

Answer: The specific heat of the metal in $1.34J/g^0C$

Explanation:

$Q_{absorbed}=Q_{released}$

As we know that,  

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

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$      

where,

$m_1$ = mass of metal = 31.7 g

$m_2$ = mass of water = 100.0 g

$T_{final}$ = final temperature = $24.6^oC$

$T_1$ = temperature of metal = $96.5^oC$

$T_2$ = temperature of water = $17.3^oC$

$c_1$ = specific heat of metal= ?

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

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

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

$-(31.7\times c_1\times (24.6-96.5)^0C)=(100.0\times 4.184\times (24.6-17.3)]$

$c_1=1.34J/g^0C$

Therefore, the specific heat of the metal in $1.34J/g^0C$