Question

If 50 g of lead (of specific heat 0.11 kcal/kg ∙ C°) at 100°C is put into 75 g of water (of specific heat 1.0 kcal/kg ∙ C°) at 0°C. What is the final temperature of the mixture?

Answer 1

Answer: The final temperature is $279.8K$

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 lead = 50 g

$m_2$ = mass of water = 75 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of lead = $100^oC=373K$

$T_2$ = temperature of water = $0^oC=273K$

$c_1$ = specific heat of lead = $0.11kcal/kg^0C$

$c_2$ = specific heat of water= $1.0kcal/kg^0C$

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

$50\times 0.11\times (T_{final}-373)=-[75\times 1.0\times (T_{final}-273)]$

$T_{final}=279.8K$

Therefore, the final temperature of the mixture will be 279.8 K.