A 30 g sample of water at 280 K is mixed with 50 g of water at 330 K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings.
heat lost by warm water + heat gained by cool water = 0
[mass warm water x specific heat H2O x (Tfinal-Tinitial)] + [mass cool water x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tf.
To calculate the final temperature of the mixture, we need to use the principle of conservation of energy, specifically the principle of heat exchange.
The principle states that the heat gained by one object is equal to the heat lost by the other object, assuming no heat loss to the surroundings.
In this case, we have two objects: one is the 30 g of water at 280 K, and the other is the 50 g of water at 330 K. We'll call the final temperature of the mixture T.
To calculate the final temperature, we can use the equation:
Heat gained by 30 g of water = Heat lost by 50 g of water
To calculate the heat gained or lost by an object, we can use the formula:
Heat gained/lost = mass * specific heat capacity * change in temperature
The specific heat capacity of water is approximately 4.18 J/g°C.
For the 30 g of water at 280 K, the change in temperature is T - 280.
So, the heat gained by the 30 g of water is:
Q1 = 30 * 4.18 * (T - 280)
For the 50 g of water at 330 K, the change in temperature is 330 - T.
So, the heat lost by the 50 g of water is:
Q2 = 50 * 4.18 * (330 - T)
Since there is no heat loss to the surroundings, Q1 is equal to Q2:
30 * 4.18 * (T - 280) = 50 * 4.18 * (330 - T)
Let's solve this equation to find the final temperature T.
First, let's simplify the equation:
30 * (T - 280) = 50 * (330 - T)
Next, distribute and combine like terms:
30T - 8400 = 16500 - 50T
Combine the T terms:
80T = 24900
Divide both sides by 80:
T = 311.25
Therefore, the final temperature of the mixture is approximately 311.25 K.
To calculate the final temperature of the mixture, we can use the principle of energy conservation.
The amount of 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 (in Joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
For water, the specific heat capacity is approximately 4.18 J/g°C.
Let's calculate the heat gained or lost by each sample of water.
For the 30 g sample of water at 280 K:
Q1 = (30 g) * (4.18 J/g°C) * (Tf - 280 K)
For the 50 g sample of water at 330 K:
Q2 = (50 g) * (4.18 J/g°C) * (Tf - 330 K)
Since there is no heat loss to the surroundings, the total heat gained by the first sample is equal to the total heat lost by the second sample. Therefore, we can set Q1 equal to -Q2:
(30 g) * (4.18 J/g°C) * (Tf - 280 K) = -(50 g) * (4.18 J/g°C) * (Tf - 330 K)
Now, let's solve for Tf.
(30 g) * (4.18 J/g°C) * Tf - (30 g) * (4.18 J/g°C) * 280 K = -(50 g) * (4.18 J/g°C) * Tf + (50 g) * (4.18 J/g°C) * 330 K
Simplifying the equation:
(30 g) * (4.18 J/g°C) * Tf - 30 * (4.18 J/g°C) * 280 = -(50 g) * (4.18 J/g°C) * Tf + 50 * (4.18 J/g°C) * 330
(125.4 J/°C) * Tf - 31392 J = -(209 J/°C) * Tf + 68670 J
Combining like terms:
(125.4 J/°C + 209 J/°C) * Tf = 68670 J + 31392 J
334.4 J/°C * Tf = 100062 J
Dividing both sides by 334.4 J/°C:
Tf = 299.1 K
Therefore, the final temperature of the mixture is 299.1 K.