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?

1 Answer
Oct 10, 2016

Answer:

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#