Question

I need help on this question

79 g of water at 95 degree Celsius were added to tap water at 22 degree Celsius, and the Temperature of the mixture after all thermal energy transfer ended at 36 degree Celsius. What was the mass of the tap water?

(teacher's answer key : 412 grams)

Im not sure what I did wrong, but anyways here's my work

MCT=-MCT

Convert grams to KeloGrams
(0.079)(4.2x10^3)(95-36)=-(M)(4.2x10^3)(22-36)

get rid of (4.2x10^3) from both sides because it is the same. So...

(0.079)(95-36)=-(M)(22-36)
4.661=-(M)(-14)
4.661/-14 =-(M)
-0.33= -M
0.33 = M

This was apparently wrong.....where did I go wrong any help please?

Answer 1

I don't see an error. Lets do it my way:
The sum heats gained is zero.
79*c*(95-36)+M*c*(22-36)=0
M= 79g*59/14=333 grams.

Answer 2

The heat lost by the hot water equals the heat gained by the tap water. The hot water cools 59 C and the tap water warms up by 14 C.
79*59*C =M*14*C

The specific heat C cancels out.

M = (59/14)*79 grams
= 333 g = 0.333 kg

I agree with your answer.

Answer 3

okay thanks alot guyz!