A person tries to heat up her bath water by adding 5.0 L of water at 80 degrees celcious to 60 L of water at 30 degrees celcious. What is the final temperature of the water?
43.33 degrees Celsius.
thanks!
To find the final temperature of the water after mixing, we can use the principle of heat transfer:
First, we need to calculate the heat gained or lost by each amount of water:
The heat gained by the 5.0 L of water can be calculated using the formula:
Q1 = m1 * c * ΔT1
Where:
Q1 = heat gained by the 5.0 L of water
m1 = mass of the 5.0 L of water
c = specific heat capacity of water (4.18 J/g°C)
ΔT1 = change in temperature for the 5.0 L of water
Since density (d) of water is 1.0 g/mL, we can assume that 1 mL of water is equal to 1 g. Therefore, the mass (m1) of the 5.0 L of water is:
m1 = V1 * d
Where:
m1 = mass of the 5.0 L of water
V1 = volume of the 5.0 L of water
d = density of water (1.0 g/mL)
Now, we can calculate the heat gained by the 5.0 L of water:
Q1 = m1 * c * ΔT1
Next, we calculate the heat gained by the 60 L of water using the same formula:
Q2 = m2 * c * ΔT2
Where:
Q2 = heat gained by the 60 L of water
m2 = mass of the 60 L of water
ΔT2 = change in temperature for the 60 L of water
Again, the mass (m2) of the 60 L of water can be calculated using the density of water:
m2 = V2 * d
Where:
m2 = mass of the 60 L of water
V2 = volume of the 60 L of water
d = density of water (1.0 g/mL)
Now, we can calculate the heat gained by the 60 L of water:
Q2 = m2 * c * ΔT2
Since heat is transferred between the two amounts of water, we can assume that the heat lost by one amount of water is equal to the heat gained by the other amount of water:
Q1 = Q2
Therefore:
m1 * c * ΔT1 = m2 * c * ΔT2
Now, let's plug in the given values to calculate the final temperature:
m1 = (5.0 L) * (1.0 g/mL)
= 5000 g
m2 = (60 L) * (1.0 g/mL)
= 60000 g
c = 4.18 J/g°C (specific heat capacity of water)
ΔT1 = 80 °C - final temperature
ΔT2 = final temperature - 30 °C
Plugging in these values, we get:
5000 g * 4.18 J/g°C * (80 °C - final temperature) = 60000 g * 4.18 J/g°C * (final temperature - 30 °C)
Simplifying the equation:
5000 * (80 - final temperature) = 60000 * (final temperature - 30)
Now, we can solve for the final temperature:
5000 * 80 - 5000 * final temperature = 60000 * final temperature - 60000 * 30
400000 - 5000 * final temperature = 60000 * final temperature - 1800000
70000 * final temperature = 1800000 + 400000
70000 * final temperature = 2200000
final temperature = 2200000 / 70000
= 31.43 °C
Therefore, the final temperature of the water is approximately 31.43 °C.
To find the final temperature of the water, we can use the principle of conservation of energy. The total heat gained by the cold water must equal the total heat lost by the hot water, assuming no heat is lost to the surroundings.
We can use the formula:
Heat gained = Heat lost
The heat gained by the cold water can be calculated by multiplying its mass (60 L) by its specific heat capacity (c) and the change in temperature (ΔT).
The heat lost by the hot water can be calculated by multiplying its mass (5.0 L) by its specific heat capacity (c) and the change in temperature (ΔT).
The specific heat capacity of water is about 4.18 J/g°C.
Since the specific heat capacity is the same for both parts of the water, we can set up the equation:
(60 L) * c * (final temperature - 30°C) = (5.0 L) * c * (80°C - final temperature)
To solve for the final temperature, we can rearrange the equation and solve for it algebraically:
(60 L * c * final temperature) - (60 L * c * 30°C) = (5.0 L * c * 80°C) - (5.0 L * c * final temperature)
From here, we simplify and solve for the final temperature:
60 L * c * final temperature + 5.0 L * c * final temperature = 5.0 L * c * 80°C + 60 L * c * 30°C
65 L * c * final temperature = 400 L * c
Divide both sides of the equation by (65 L * c):
final temperature = (400 L * c) / (65 L * c)
final temperature ≈ 6.15°C
Therefore, the final temperature of the water is approximately 6.15 degrees Celsius.