How much water at 80.0C must be added to 200.g of water at 10.0C to give a final temperature of 36.0cc?
I know the answer comes to 1.2x10^-1kg but I need to know how to solve the question... ie: the long work involved.
To solve this question, we can use the principle of conservation of energy, specifically the equation for heat transfer:
q1 + q2 = 0
where q1 is the heat gained by the colder water (200 g) and q2 is the heat lost by the hotter water.
First, let's calculate the heat gained by the colder water:
q1 = m1 * c1 * ΔT1
where:
m1 is the mass of the cold water (200 g),
c1 is the specific heat capacity of water (4.186 J/g°C),
ΔT1 is the change in temperature of the cold water (final temperature - initial temperature).
ΔT1 = 36.0°C - 10.0°C = 26.0°C
q1 = (200 g) * (4.186 J/g°C) * (26.0°C)
q1 = 21772 J
Now, let's calculate the heat lost by the hotter water:
q2 = m2 * c2 * ΔT2
where:
m2 is the mass of the hot water we want to find,
c2 is the specific heat capacity of water (4.186 J/g°C),
ΔT2 is the change in temperature of the hot water (final temperature - initial temperature).
ΔT2 = 36.0°C - 80.0°C = -44.0°C (negative value since the hot water is losing heat)
q2 = (m2) * (4.186 J/g°C) * (-44.0°C)
q2 = -184.504 m2 J
Since q1 + q2 = 0, we can set up the equation:
21772 J + (-184.504 m2 J) = 0
Solving for m2:
184.504 m2 = 21772
m2 = 21772 / 184.504
m2 = 117.9 g
Therefore, approximately 117.9 g (or 0.1179 kg, rounded to 3 significant figures) of water at 80.0°C must be added to 200 g of water at 10.0°C to give a final temperature of 36.0°C.