Question

What is the final temperature when 100 g of water at 25`"^o`C is mixed with 75 g of water at 50`"^o`C?

Answer 1

You must remeber than total energy remains constant: the thermic energy that wins the cold water must be equal to the thermic energy that loses the hot one.

Explanation:

The heat exchanging of a system with the outside when a mass `m` of a material having a specific heat `C_e` undergoes a temperature variation `Delta t` is

`Q = m cdot C_e cdot Delta t`

Then, if we mix a mass `m_1` of water (specific heat `C_e (H_{2}O) = 4,18 kJ cdot kg^{- 1} cdot ºC^{- 1} )` at initial temperature `t_1` with a second quantity of water, with mass `m_2` and temperature `t_2`, the hot water will cool while the cold will warm until both reach the same temperature. Asume `t_1 gt t_2` and suppose that the final temperature of the system is t. Then:

`m_1 cdot C_e (H_{2}O) cdot (t_1 - t) = m_2 cdot C_e (H_{2}O) cdot (t - t_2)`

Substituting the values we have:

`75 cdot (50 - t) = 100 cdot (t - 25) rArr 3750 - 75 t = 100 t - 2500 rArr t = 35,7ºC`