how many ml of water at 10 degrees celsius must be added to 75 ml of water at 35 degrees celsius to obtain a final temperature of 19 degrees celsius?
[mL1 x specific heat x (T1final-T1inital)] + [mL2 x specific heat x (T2final-T2initial) = 0
To solve this question, we can use the principle of conservation of energy. The heat gained by the cold water is equal to the heat lost by the hot water when they mix together.
We can use the formula for heat transfer:
Q = mcΔT
Where:
Q = heat transfer
m = mass of the water
c = specific heat capacity of water (which is approximately 4.18 J/g·°C)
ΔT = change in temperature
First, let's calculate the heat gained by the cold water:
Q1 = mcΔT
Substituting the values we know:
Q1 = m1 * 4.18 * (19 - 10)
Q1 = m1 * 4.18 * 9
Next, let's calculate the heat lost by the hot water:
Q2 = mcΔT
Substituting the values we know:
Q2 = m2 * 4.18 * (35 - 19)
Q2 = m2 * 4.18 * 16
Since the heat gained by the cold water is equal to the heat lost by the hot water, we have:
Q1 = Q2
m1 * 4.18 * 9 = m2 * 4.18 * 16
Now we can calculate the mass of the cold water that needs to be added.
Let's say the mass of the cold water is m1 and the mass of the hot water is m2.
m1 * 9 = m2 * 16
We know the mass of the hot water (75 ml), so let's substitute it in:
m1 * 9 = 75 * 16
Now we can solve for m1:
m1 = (75 * 16) / 9
m1 ≈ 133.33 ml
Therefore, approximately 133.33 ml of water at 10 degrees Celsius must be added to 75 ml of water at 35 degrees Celsius to obtain a final temperature of 19 degrees Celsius.