Question

When a 40.0 g sample of a metal at 25.00 °C is added to 65.0 g of water at 100.00 °C, the final temperature of both the water and metal is 93.27 °C. The specific heat of water is 4.184 J/(g×C). What is the specific heat of the metal?

Answer 1

Answer:  Specific heat of the metal is $0.670J/g^0C$

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 metal = 40.0 g

$m_2$ = mass of water = 65.0 g

$T_{final}$ = final temperature = $93.27^oC=(93.27+273)K=366.27K$

$T_1$ = temperature of metal = $25^oC=(25+273)K=298K$

$T_2$ = temperature of water = $100^oC=(100+273)K=373K$

$c_1$ = 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

$40.0\times c_1\times (366.27-298)=-[65.0\times 4.184\times (366.27-373)]$

$c_1=0.670$

Therefore, the specific heat of the metal is $0.670J/g^0C$