When 86.7 g of water at a temperature of 73 °C is mixed with an unknown mass of water at a temperature of 22.3 °C the final temperature of the resulting mixture is 61.7 °C. What was the mass of the second sample of water?
(Specific heat of water 4.184 J·g-1·°C-1)
To solve this question, we can use the principle of conservation of energy, which states that the heat gained by one substance is equal to the heat lost by another substance in a closed system.
First, we need to calculate the heat gained by the cooler water, which can be calculated using the formula:
Q1 = m1 * c * (Tf - Ti)
Where:
Q1 is the heat gained by the cooler water
m1 is the mass of the cooler water (86.7 g)
c is the specific heat capacity of water (4.184 J·g-1·°C-1)
Tf is the final temperature (61.7 °C)
Ti is the initial temperature of the cooler water (22.3 °C)
Plugging in the values, we get:
Q1 = 86.7 g * 4.184 J·g-1·°C-1 * (61.7 °C - 22.3 °C)
Next, we want to find the mass of the second sample of water, which we can calculate using the equation:
Q1 = Q2
Where:
Q1 is the heat gained by the cooler water (calculated above)
Q2 is the heat lost by the warmer water
Since the final temperature is the same for both substances, the heat lost by the warmer water can be calculated using:
Q2 = m2 * c * (Tf - Ti)
Where:
m2 is the mass of the second sample of water (what we want to find)
c is the specific heat capacity of water (4.184 J·g-1·°C-1)
Tf is the final temperature (61.7 °C)
Ti is the initial temperature of the warmer water (73 °C)
Now we can set up the equation:
Q1 = Q2
86.7 g * 4.184 J·g-1·°C-1 * (61.7 °C - 22.3 °C) = m2 * 4.184 J·g-1·°C-1 * (61.7 °C - 73 °C)
Simplifying the equation and solving for m2:
(86.7 g * (61.7 °C - 22.3 °C)) / (61.7 °C - 73 °C) = m2
Calculating this expression will give us the mass of the second sample of water.