Question

If 500g of water at 20 degrees C is mixed with 750g of water at 30 degrees C, what will the temperature of the mixture be?

Answer 1

Answer:

$26 ^\circ C$

Explanation:

Given that the temperature of  $500g$ of water and $750 g$ of water are

at $20^{\circ}C$ and $30^\circ C$ respectively.

Let $m_1=500g$, $T_1= 20^\circ C$

and $m_2=750g$, $T_2= 30^\circC$.

The specific heat capacity of water is,

$C= 4.186 J/g ^\circ C$.

Let the final temperature of the mixture be $T^\circ C$.

As there is no energy loss, so, the energy loss by the water at higher temperature, i.e. $30^\circ C$, will be equal to the energy gain by the water at lower temperature, i.e. $20^{\circ}C$.

$m_2C (T_2-T)=m_1C(T-T_1)$

$\Rightarrow m_2 (T_2-T)=m_1(T-T_1)$ [ both sides divided by $C$ ]

$\Rightarrow m_2T_2-m_2T=m_1T-m_1T_1$

$\Rightarrow m_1T+m_2T=m_1T_1+m_2T_2$

$\Rightarrow T=\frac{m_1T_1+m_2T_2}{m_1+m_2}$

Now, putting the given value in the above equation, we have

$\Rightarrow T=\frac {500\times 20+750\times 30}{500+750}$

$\Rightarrow T=26^\circ C.$

Hence, the temperature of the mixture will be $26 ^\circ C$.