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

 P: 17 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?
HW Helper
P: 6,678
 Quote by Axoren 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?
Total Q = 0 .

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

AM
 P: 17 Thank you very much! I don't know how I missed that.
 P: 2 Heat and Internal Energy and Heat and Temperature Change: Specific Heat Capacity 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? Please help.
 P: 2 Just kidding i'm stupid. i figured it out... sorry for wasting your time.

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