You KNOW that the cooler water will cool some of the warmer water to give a final temperature somewhere between 25 and 65. What will be the final temperature? Who Knows? There is no information about HOW MUCH OF THE WARMER WATER was used. IF you know how much the final temperature can be calculated this way.
[mass cool water x specific heat water x (Tfinal - Tinitial)] + [mass warm water x specific heat water x (Tfinal - Tinitial)] = 0
[25 x 4.18 x (Tf - 25)] + [mass warm water x 4.18 x (Tfinal - 65)] = 0
Plug in the mass of the warm water and solve for Tf (final T). Post your work if you get stuck.
If you mix 25 grams of water at 25 Degrees with a larger sample of water at 65 Degrees what will be the final temperature and why
2 answers
mass * specific heat * change in temp = heat in or out
heat in = 25 * sh * (T-25)
heat out = m * sh *(65 - T)
25 T - 625 = 65 m - m T
(25 + m) T= 65 m + 625
obviously you need m , the mass of the hot water, to find T
heat in = 25 * sh * (T-25)
heat out = m * sh *(65 - T)
25 T - 625 = 65 m - m T
(25 + m) T= 65 m + 625
obviously you need m , the mass of the hot water, to find T