Question

10 gm of water at 10°C is mixed with 20 gm of
water at 30°C. The final temperature of the
mixture will be​

Answer 1

Answer: $23.33^0C$

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 water which absorbed heat= 10 g

$m_2$ = mass of water which released heat= 20 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of water which absorbed heat = $10^oC$

$T_2$ = temperature of water which released heat= $30^oC$

$c_1$ = specific heat of water= $1.0kcal/kg^0C$

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

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

$10\times 1.0\times (T_{final}-10)=-[20\times 1.0\times (T_{final}-30)]$

$T_{final}=23.3^0C$

Thus final temperature of the  mixture will be​ $23.33^0C$

Answer 2

Explanation:

its correct I hope it's helpful