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

`26 ^\circ C`

Step-by-step 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=(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* 20+750* 30}{500+750}`

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

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