(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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

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