Question

25 g of 116oC steam are bubbled into 0.2384 kg of water at 8oC. Find the final temperature of the system.

Answer 1

Answer: The final temperature of the system will be $13.14^0C$

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 steam = 25 g

$m_2$ = mass of water = 0.2384 kg = 238.4 g   (1kg=1000g)

$T_{final}$ = final temperature = ?

$T_1$ = temperature of steam = $116^oC$

$T_2$ = temperature of water = $8^oC$

$c_1$ = specific heat of steam = $1.996J/g^0C$

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

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

$25g\times 1.996J/g^0C\times (T_{final}-116)=-[238.4g\times 4.184J/g^0C\times (T_{final}-8)]$

$T_{final}=13.14^0C$

Therefore, the final temperature of the system will be $13.14^0C$