# Two 10.0 L buckets of hot water at 65 degrees celsius are mixed with one 10.0 L bucket of cold...

## Question:

Two 10.0 L buckets of hot water at 65 degrees celsius are mixed with one 10.0 L bucket of cold water at 18 degrees celsius. Calcualte the final equilibrium temperature after mixing, assuming no thermal energy loss to the environment.

## Equilibrium temperature

When two liquid at different temperatures are mixed, then the equilibrium temperature of the mixture can be calculated by equating the heat lost by the liquid of higher temperature to heat gained by the liquid of lower temperature.

We are given:

• The temperature of the hot water is {eq}T_1\ =\ 65^\circ C{/eq}
• The volume of the hot water is {eq}V_1\ =\ 10.0\ \rm L{/eq}
• The temperature of the cold water is {eq}T_2\ =\ 18^\circ C{/eq}
• The volume of the cold water is {eq}v_2\ =\ 10.0\ \rm L{/eq}

Let,

• The equilibrium temperature of the mixture is T.
• The specific heat capacity of the water is c.
• The density of the water is {eq}\rho {/eq}

Since the hot water is mixed with the cold water, so some part of the heat of the hot water will transfer to the cold water to raise the temperature up to its equilibrium temperature.

Now,

From conservation of energy:

{eq}\begin{align} Energy\ lost\ by\ hot\ water\ &=\ Energy\ gained\ by\ cold\ water\\ m_1c(T_1-T)\ &=\ m_2c(T-T_2)\\[0.3 cm] \rho_wV_1c(T_1-T)\ &=\ \rho_wV_2c(T-T_2)\\[0.3 cm] V_1(T_1-T)\ &=\ V_2(T-T_2)\\[0.3 cm] 10.0\times 10^{-3}\ \rm m^3(65^\circ C-T)\ &=\ 10.0\times 10^{-3}\ \rm m^3(T-18^\circ C)\\[0.3 cm] 650^\circ C-10T\ &=\ 10T-180^\circ C\\[0.3 cm] 20\ T\ &=\ 650^\circ C-180^\circ C\\[0.3 cm] T\ &=\ \dfrac{470^\circ C}{20}\\ &=\ \boxed{\color{green}{23.5^\circ C}}\\ \end{align} {/eq}