# Heat and Internal Energy and Heat and Temperature Change: Specific Heat Capacity

1. Nov 9, 2008

### Axoren

1. The problem statement, all variables and given/known data
When you take a bath, how many kilograms of hot water (49°C) must you mix with cold water (12°C) so that the temperature of the bath is 36°C? The total mass of water (hot plus cold) is 191 kg. Ignore any heat flow between the water and its external surroundings.

Temperature of the Cold Water: 12 C
Temperature of the Hot Water: 49 C
Temperature of the Total Water: 36 C
Mass of the Cold Water: y, 191-x
Mass of the Hotwater: x, 191-y
Mass of the Total Water: 191 kg

2. Relevant equations
Q =mcΔT
Where Q is Joules of Energy, m is Mass of the substance (kg), c is the specific heat coefficient, ΔT is change in Temperature (in Celsius)

3. The attempt at a solution
Total Q = 191 kg * 4186 J/kgC * 36 C
Total Q = 28782936 J
Hot Q = X kg * 4186 J/kgC * 13 C
Cold Q = (191 - X) kg * 4186 J/kgC * 24 C
28782936 J = Hot Q + Cold Q
28782936 J = (X kg * 4186 J/kgC * 13 C) + ((191 - X) kg * 4186 J/kgC * 24 C)
28782936 J = (X kg * 54418 J/kg) + ((191 - X) kg * 100464 J/kg)
28782936 J = 54418X J + 19188624 J - 100464X J
28782936 J = 19188624 J - 46046X J
9594312 J = 46046X J
208.36 kg = X (Answer only needs to be to 2 decimal places)

^This does not make sense, because the mass of the hot water is larger than the mass of the total water. Assistance?

Last edited: Nov 9, 2008
2. Nov 9, 2008

### Andrew Mason

Total Q = 0 .

So the equation should be $Q_h + Q_c=0$

AM

3. Nov 9, 2008

### Axoren

Thank you very much! I don't know how I missed that.

4. Feb 12, 2011

### ahef7

Hey I really need this answered I've been working on it for 3 hours help.
Question: You want to take a bath with the water temperature at 35.0 degrees C. The water temperature is 38 degrees C from the hot water tap and 11 degrees C from the cold water tap. You fill the tub with a total of 187 kg of water. How many kilograms of water from the hot water tap do you use?