Question

4 OT 5

Answer 1

Answer: The specific heat capacity of metal is $1.31J/g^0C$

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 = 35 g

$m_2$ = mass of water = 220 g

$T_{final}$ = final temperature = $30^0C$

$T_1$ = temperature of metal = $130^oC$

$T_2$ = temperature of water = $25^oC$

$c_1$ = 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

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

$35\times c_1\times (30-130)^0C=-[220g\times 4.184\times (30-25)]$

$c_1=1.31J/g^0C$

Therefore, the specific heat capacity of metal is $1.31J/g^0C$