You have a mass of hot water at 90 C and twice the mass of cool water at 10C. The hot and cold water are mixed together. Will the final temperature be midway between 10C and 90C, closer to 90C, or closer to 10C? Explain your reasoning.
What do you think? You started out with more cold water than hot.
If the amounts of cold and hot would have been equal, the equilibrium temperature would be 50C (halfway between 10 and 90).
that makes sense i just didn't think it could be that easy
To determine the final temperature, we need to consider the principle of energy conservation. When the hot water and cold water are mixed together, heat will transfer from the hot water to the cold water until thermal equilibrium is reached.
The amount of heat transferred from the hot water to the cold water is given by the equation:
Q = mcΔT
Where:
- Q is the amount of heat transferred
- m is the mass of water
- c is the specific heat capacity of water
- ΔT is the change in temperature
In this case, let's assume the mass of the hot water is m1 and the mass of the cold water is m2. Given that the hot water has a mass twice that of the cold water, we can state that m1 = 2m2.
Since the heat lost by the hot water is equal to the heat gained by the cold water (Q1 = Q2), we can write the following equation:
m1cΔT1 = m2cΔT2
In this scenario, ΔT1 is the final temperature minus the initial temperature of the hot water (90C - T1), and ΔT2 is the final temperature minus the initial temperature of the cold water (T2 - 10C).
Substituting m1 = 2m2, we get the equation:
2m2c(90C - T1) = m2c(T2 - 10C)
Simplifying the equation, we have:
180C - 2cT1 = cT2 - 10c
Rearranging the terms, we get:
180C + 10C = 2cT1 + cT2
Combining like terms, we have:
190C = 3c(T1 + T2)
Dividing both sides by 3c, we get:
T1 + T2 = 190C / 3c
Therefore, the sum of the final temperatures T1 + T2 is a constant value, which means the final temperature will fall somewhere between the initial temperatures of 10C and 90C, but closer to the temperature of the hot water at 90C.
To determine the final temperature when hot and cold water are mixed together, we can use the principle of conservation of energy, specifically the principle of heat exchange.
The principle of heat exchange states that when two bodies at different temperatures are put in contact with each other, heat flows from the body at a higher temperature to the body at a lower temperature until thermal equilibrium is reached. The heat lost by the hot water is equal to the heat gained by the cold water.
In this scenario, we have a mass (m1) of hot water at 90°C, and twice the mass (2m1) of cold water at 10°C. Let's assume the final temperature after mixing is Tf.
The amount of heat lost by the hot water is given by Q1 = m1 * c * (T1 - Tf), where c is the specific heat capacity of water (assumed constant) and T1 is the initial temperature of the hot water (90°C).
The amount of heat gained by the cold water is given by Q2 = (2m1) * c * (Tf - T2), where T2 is the initial temperature of the cold water (10°C).
Since the principle of heat exchange states that the two quantities of heat are equal, we can set up the equation Q1 = Q2:
m1 * c * (T1 - Tf) = (2m1) * c * (Tf - T2)
By cancelling out the mass (m1), rearranging the equation, and solving for Tf, we have:
Tf = T1 - (2(T1 - T2))
Tf = T1 - 2T1 + 2T2
Tf = 2T2 - T1
Plugging in the values, Tf = 2(10°C) - 90°C
Tf = 20°C - 90°C
Tf = -70°C
From the calculation, we can see that the final temperature is -70°C, which is closer to 10°C (the initial temperature of the cold water) rather than midway between 10°C and 90°C.
Therefore, the final temperature is closer to 10°C.