Respuesta :
Answer:
[tex]26 ^\circ C[/tex]
Explanation:
Given that the temperature of [tex]500g[/tex] of water and [tex]750 g[/tex] of water are
at [tex]20^{\circ}C[/tex] and [tex]30^\circ C[/tex] respectively.
Let [tex]m_1=500g[/tex], [tex]T_1= 20^\circ C[/tex]
and [tex]m_2=750g[/tex], [tex]T_2= 30^\circC[/tex].
The specific heat capacity of water is,
[tex]C= 4.186 J/g ^\circ C[/tex].
Let the final temperature of the mixture be [tex]T^\circ C[/tex].
As there is no energy loss, so, the energy loss by the water at higher temperature, i.e. [tex]30^\circ C[/tex], will be equal to the energy gain by the water at lower temperature, i.e. [tex]20^{\circ}C[/tex].
[tex]m_2C (T_2-T)=m_1C(T-T_1)[/tex]
[tex]\Rightarrow m_2 (T_2-T)=m_1(T-T_1)[/tex] [ both sides divided by [tex]C[/tex] ]
[tex]\Rightarrow m_2T_2-m_2T=m_1T-m_1T_1[/tex]
[tex]\Rightarrow m_1T+m_2T=m_1T_1+m_2T_2[/tex]
[tex]\Rightarrow T=\frac{m_1T_1+m_2T_2}{m_1+m_2}[/tex]
Now, putting the given value in the above equation, we have
[tex]\Rightarrow T=\frac {500\times 20+750\times 30}{500+750}[/tex]
[tex]\Rightarrow T=26^\circ C.[/tex]
Hence, the temperature of the mixture will be [tex]26 ^\circ C[/tex].
