If 100 g of water (with a specific heat capacity of 4200 J/kg°C) at 10oC is mixed with 100 g of another liquid (with a specific heat capacity of 2100 J/kg°C ) at 70oC are mixed, what would you expect to get for the final temperature?

We can find the final temperature of the mixture of 100 g of water and 100 g of liquid by using the Black Principle, which states that "the heat released by a higher temperature substance is equal to the heat absorbed by a lower temperature substance". Therefore:

[tex]\boxed{Q_{released}=Q_{absorbed}}[/tex]

and the formula for heat:

[tex]\boxed{Q=mc\Delta T}[/tex]

where:

[tex]Q=\texttt{heat}[/tex]
[tex]m=\texttt{mass}[/tex]
[tex]c=\texttt{heat capacity}[/tex]
[tex]\Delta T=\texttt{temperature difference}[/tex]

Given:

[tex]m_{w}=100\ g=0.1\ kg[/tex]
[tex]c_w=4200\ J/kg^oC[/tex]
[tex]T_0_w=10^oC[/tex]
[tex]m_{l}=100\ g=0.1\ kg[/tex]
[tex]c_l=2100\ J/kg^oC[/tex]
[tex]T_0_l=70^oC[/tex]

Then:

[tex]\begin{aligned}Q_{released}&=Q_{absorbed}\\Q_l&=Q_w\\m_l\cdot c_l(T_0_l-T)&=m_w\cdot c_w(T-T_0_l)\\0.1(2100)(70-T)&=0.1(4200)(T-10)\\70-T&=2(T-10)\\2T+T&=70+20\\3T&=90\\T&=\bf 30^oC\end{aligned}[/tex]