Steam of mass 11.0 g at 100°C is added to 114.8 g of ice at 0.00°C. Find the final temperature of the system.
The answer is 0
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To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the steam when it condenses to water will be equal to the heat gained by the ice when it melts.
First, let's find the heat lost by the steam. We can use the formula:
Q = m * L
where Q is the heat lost, m is the mass, and L is the heat of condensation (latent heat) of water, which is 2.26 ×10^6 J/kg.
The mass of the steam, m1 = 11.0 g = 0.011 kg.
So, the heat lost by the steam, Q1 = m1 * L.
Next, let's find the heat gained by the ice. We'll use the formula:
Q = m * L
where Q is the heat gained, m is the mass, and L is the heat of fusion (latent heat) of ice, which is 3.34 ×10^5 J/kg.
The mass of the ice, m2 = 114.8 g = 0.1148 kg.
So, the heat gained by the ice, Q2 = m2 * L.
Since the heat lost by the steam is equal to the heat gained by the ice, we have:
Q1 = Q2
m1 * L = m2 * L
0.011 kg * 2.26 ×10^6 J/kg = 0.1148 kg * 3.34 ×10^5 J/kg
Now, we can solve for the final temperature of the system by using the formula:
Q = m * c * ΔT
where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Assuming the specific heat capacity of water is 4186 J/(kg·°C), we can calculate the final temperature using the equation:
Q1 = m1 * c * (T2 - Tf)
where Tf is the final temperature.
Combining the equations, we have:
m1 * L = m1 * c * (T2 - Tf)
0.011 kg * 2.26 ×10^6 J/kg = 0.011 kg * 4186 J/(kg·°C) * (100°C - Tf)
Simplifying the equation, we get:
2.26 ×10^6 J = 4186 J/(kg·°C) * 110°C - 4186 J/(kg·°C) * Tf
Now, let's solve for Tf:
2.26 ×10^6 J = 460,460 J - 4186 J/(kg·°C) * Tf
2.26 ×10^6 J - 460,460 J = -4186 J/(kg·°C) * Tf
1.79954 ×10^6 J = 4186 J/(kg·°C) * Tf
Tf = 1.79954 ×10^6 J / (4186 J/(kg·°C))
Tf = 429.83 °C
Therefore, the final temperature of the system is approximately 429.83 °C.