two hundred fifty grams of water at 80 degree celsius is poured into a styrofoam cap of negligible heat capacity containing 180g of water at 10 degree celsius. after an additional 300 g of water to the cap, the mixture comes to the equilibrium temperature of 30 degree celsius. what was the temperature of the additional 300 g of water?
To find the temperature of the additional 300g of water, we can use the principle of conservation of energy.
First, let's find the amount of heat gained or lost by each component of the system. We'll consider the water at 80°C, the water at 10°C, and the additional 300g of water.
The heat gained or lost by each component can be calculated using the formula:
Q = mcΔT,
where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
For the water at 80°C:
Q1 = 250g * c * (30°C - 80°C),
where c is the specific heat capacity of water.
For the water at 10°C:
Q2 = 180g * c * (30°C - 10°C).
For the additional 300g of water:
Q3 = 300g * c * (30°C - x),
where x is the temperature we want to find.
Since the system is in equilibrium, the heat gained by the water at 80°C and the heat gained by the water at 10°C should be equal to the heat gained by the additional 300g of water. Therefore, we can set up the following equation:
Q1 + Q2 = Q3,
250g * c * (30°C - 80°C) + 180g * c * (30°C - 10°C) = 300g * c * (30°C - x).
Now we can solve this equation for x to find the temperature of the additional 300g of water.
To solve this problem, we can use the principle of conservation of energy, which states that the total heat gained by one substance must equal the total heat lost by the other substance.
Let's break down the steps:
Step 1: Calculate the heat lost by the initial 180g of water.
The formula to calculate the heat lost or gained is given by Q = mcΔT, where:
- Q is the heat lost or gained
- m is the mass of the substance
- c is the specific heat capacity of water (approximately 4.18 J/g°C)
- ΔT is the change in temperature
The mass of the initial 180g of water is m1 = 180g.
The initial temperature of the 180g of water is T1 = 10°C.
The final temperature of the 180g of water is T_final = 30°C.
So, the heat lost by the initial 180g of water is:
Q1 = m1 * c * (T_final - T1)
= 180g * 4.18 J/g°C * (30°C - 10°C)
= 180g * 4.18 J/g°C * 20°C
= 15048 J
Step 2: Calculate the heat lost by the 250g of water initially at 80°C.
The mass of the 250g of water is m2 = 250g.
The initial temperature of the 250g of water is T2 = 80°C.
The final temperature of the 250g of water is T_final = 30°C.
So, the heat lost by the 250g of water initially at 80°C is:
Q2 = m2 * c * (T_final - T2)
= 250g * 4.18 J/g°C * (30°C - 80°C)
= 250g * 4.18 J/g°C * (-50°C)
= -52350 J (since Q2 is negative)
Step 3: Calculate the heat gained by the additional 300g of water.
The mass of the additional 300g of water is m3 = 300g.
The initial temperature of the additional 300g of water is T3 (unknown).
The final temperature of the 300g of water is T_final = 30°C.
Let's assume the initial temperature of the additional 300g of water is T3.
So, the heat gained by the additional 300g of water is:
Q3 = m3 * c * (T_final - T3)
= 300g * 4.18 J/g°C * (30°C - T3)
Step 4: Apply the principle of conservation of energy.
According to the principle of conservation of energy, the total heat gained by one substance (in this case, the additional 300g of water) must be equal to the total heat lost by the other two substances (the 180g and 250g of water).
Q3 = -Q1 - Q2
300g * 4.18 J/g°C * (30°C - T3) = -15048 J - (-52350 J)
1254 J/g°C * (30°C - T3) = 52350 J - 15048 J
37620 J - 1254 J * T3 = 37302 J
Now, we solve for T3:
-1254 J * T3 = 37302 J - 37620 J
-1254 J * T3 = -318 J
T3 = -318 J / -1254 J
T3 ≈ 0.254°C
Therefore, the temperature of the additional 300g of water is approximately 0.254°C.