Wha is the final temperature of this mixture?

Question

I'm reviewing some practice thermodynamics questions for my upcoming physics exam but I can't figure out how to set these up properly.

 

1. If you mix two parts hot 80° C water with one part 10° C cool water, what is the temperature of the mixture in °C?

 

2. You add 20° C water to a 90° C metal bowl. There is 10 kg of water and 10 kg of metal. Find the temperature of the mixture in °C. c = 0.1 cal/g/K

 

I tried mCΔT=mCΔT for both but I'm not getting a solution because the masses and the specific heats cancel out and I end up without a way to solve for the final temperature. Thank you in advance for any assistance. 

Answer 1

For part 1. ,   The heat lost by the 80° water equals the heat gained by the 10° water.

 

The problem talks about two parts and one part.   To make this specific, take the mass of the cool water to be 100 g  and the mass of the hot water to be 200 g.   Then heat lost = heat gained can be written as

 

      200 x  1 x (80 - T_f) =  100 x 1 x (T_f -10)

  where T_f is the final temperature and the 1's are there because the specific heat of water is  1 c/g/deg

 

  This equation can be rearranged in algebra to give   170 = 3 T_f   so   T_f = 56.67

      Notice that the 100 g and 200 grams did not come in except in the 2 to 1 ratio.

 

For part 2.  , The heat lost by the metal must equal the heat gained by the water.

  The heat lost by  the metal is 10 000 x .1 x (90 - T_f)    the heat gained by the water is

    10 000 x 1 x (T_f - 20)   .   Setting this equal leads to 29 = 1.1T_f  so  T_f =  26.4

  The .1 is there   because it is the specific heat of the metal.