# If 200g of water is at 80.0C, how much water at temperature of 25.0C will need to be added in order to bring the final temperature of the mixture to 60.0C? (200)(1.0)(80-f) = (x)(1.0)(f-25) 16000-200f = xf-25x 16000+25x = xf+200f I'm confused with this problem.

It looks like you're trying to solve for two unknowns. Since you know the final temperature and the temperatures of the hot and cold water before mixing, you can replace 80-f and f-25 with the actual changes in temperature and solve for x, the mass of cold water.

You've started out showing that the heat lost by the warm water equals the heat gained by the cold water, which is correct. For both the warm and cold water the heat change equals mc `Delta` T, where m is the mass of the water, c is the specific heat capacity and `Delta`  T is the change in temperature. The value of 1.0 that you used is the specific heat capacity of water in units of calories/gram-degree C. Here's what the equation should look like:

mc`Delta` T (warm water) = mc `Delta` T (cold water)

(200g)(1.0 cal/gram-*C)(80.0-60.0 *C)=(x)(1.00 cal/gram-*C)(60.0--25.0*C)

4000 cal = 35x cal/g

x= 4000/35 g = 114 g

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