Calculate the temperature found when 10kg of water at 100°C is mixed with 20kg of water at 40°C.
say specific heat of water = k (does not matter here)
10 k (100 - T) = 20 k (T - 40)
1000 - 10 T = 20 T - 800
1800 = 30 T
T = 60 deg C
60 deg c
Well, as a master of clowny calculations, I can certainly give it a shot! But brace yourself for some clown science.
So, let's call the final temperature "XTreme Fahrenheit" (yes, I like to make things extreme). To find it, we can unleash the power of clown thermodynamics!
First, we take the temperature difference between the two water masses: 100°C - 40°C = 60°C.
Next, we need to account for the masses of the two water quantities. The ratio between the masses is 10kg : 20kg, or in clown terms, 1 clown car : 2 clown cars.
Since the 20kg water is double the mass of the 10kg water, it's like having two clown cars jumping into the pool with a splash!
Now, let's imagine the two clown cars splashing into the water and creating some commotion. In the end, their "XTreme Fahrenheit" temperature reaches a consensus.
So, we divide the temperature difference (60°C) by the ratio of masses (2 clown cars) -> 60°C / 2 = 30°C.
Therefore, the "XTreme Fahrenheit" temperature is 30°C. Tadaaaa!
But remember, this result doesn't account for any clown physics or real-world thermodynamics. So, please consult a professional for the actual answer.
To calculate the final temperature when two substances are mixed, we can use the principle of heat transfer:
Q1 = Q2
where Q1 is the heat gained by the cold substance, and Q2 is the heat lost by the hot substance.
The heat gained or lost by a substance can be calculated using the formula:
Q = m * C * ΔT
where Q is the heat gained or lost, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature.
Given:
Mass of water 1 (hot water) = 10 kg
Initial temperature of water 1 (hot water) = 100°C
Mass of water 2 (cold water) = 20 kg
Initial temperature of water 2 (cold water) = 40°C
We need to find the final temperature when the two substances are mixed.
First, let's calculate the heat gained and lost by each substance:
Heat gained by water 2:
Q1 = m1 * C * ΔT1
= 10 kg * 4.18 J/g°C * (Tf - 40°C)
Heat lost by water 1:
Q2 = m2 * C * ΔT2
= 20 kg * 4.18 J/g°C * (100°C - Tf)
Since Q1 = Q2, we can equate the equations and solve for the final temperature Tf:
10 kg * 4.18 J/g°C * (Tf - 40°C) = 20 kg * 4.18 J/g°C * (100°C - Tf)
Now let's solve for Tf:
10(Tf - 40) = 20(100 - Tf)
10Tf - 400 = 2000 - 20Tf
30Tf = 2400
Tf = 80°C
Therefore, the final temperature when 10kg of water at 100°C is mixed with 20kg of water at 40°C is 80°C.
To calculate the final temperature when two bodies of water at different temperatures are mixed together, we can use the principle of thermal equilibrium. This principle states that when two bodies are in contact with each other and no heat is being lost to the surroundings, the heat gained by one body is equal to the heat lost by the other body.
To find the final temperature, we can use the equation:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Where:
- m1 is the mass of the first body of water
- c1 is the specific heat capacity of water (4.186 J/g°C)
- ΔT1 is the change in temperature of the first body of water
- m2 is the mass of the second body of water
- c2 is the specific heat capacity of water (4.186 J/g°C)
- ΔT2 is the change in temperature of the second body of water
Let's use this equation to calculate the final temperature when 10kg of water at 100°C is mixed with 20kg of water at 40°C.
First, calculate the change in temperature for each body of water:
ΔT1 = Tf - Ti1
= Tf - 100°C
ΔT2 = Tf - Ti2
= Tf - 40°C
Next, rearrange the equation to solve for the final temperature (Tf):
(m1 * c1 * ΔT1 + m2 * c2 * ΔT2) / (m1 * c1 + m2 * c2) = Tf
Substitute the values into the equation:
(10kg * 4.186 J/g°C * (Tf - 100°C) + 20kg * 4.186 J/g°C * (Tf - 40°C)) / (10kg * 4.186 J/g°C + 20kg * 4.186 J/g°C) = Tf
Simplify the equation and solve for Tf:
(41.86 Tf - 4186 J + 83.72 Tf - 3348.8 J) / (41.86 + 83.72) = Tf
(125.58 Tf - 7534.8 J) / 125.58 = Tf
125.58 Tf - 7534.8 J = 125.58 Tf
7534.8 J = 0.58 Tf
Tf = 7534.8 J / 0.58
Tf ≈ 13014.5 J / 0.58
Tf ≈ 22455.2°C
Therefore, when 10kg of water at 100°C is mixed with 20kg of water at 40°C, the final temperature is approximately 22,455.2°C.