Determine the final temperature of a 200 g block of ice at -10°C when it is placed in a container with 400 g of water at 80°C. Calculate the total heat transfer and the nature of the final state (e.g., all ice melted).
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To determine the final temperature, we must calculate the heat transfer between the ice and the water until they reach thermal equilibrium.
First, let's calculate the heat transfer from the water to the ice. We'll use the formula:
Q = m * c * ΔT
where Q is the heat transfer in Joules, m is the mass in grams, c is the specific heat capacity in J/g°C, and ΔT is the change in temperature in °C.
The specific heat capacity of water is 4.18 J/g°C, and for ice, it is 2.09 J/g°C.
For the water:
Q1 = 400 g * 4.18 J/g°C * (Tf - 80°C)
For the ice:
Q2 = 200 g * 2.09 J/g°C * (Tf - (-10°C))
Since the total heat transfer is equal for both the water and the ice, we can set up an equation:
Q1 = Q2
400 g * 4.18 J/g°C * (Tf - 80°C) = 200 g * 2.09 J/g°C * (Tf - (-10°C))
Simplifying the equation:
(400 * 4.18)(Tf - 80) = (200 * 2.09)(Tf + 10)
1672Tf - 33440 = 418Tf + 4180
1672Tf - 418Tf = 33440 + 4180
1254Tf = 37620
Tf = 37620 / 1254
Tf ≈ 30°C
Therefore, the final temperature of the system is approximately 30°C.
To calculate the total heat transfer, we substitute the final temperature into one of the equations for Q1 or Q2:
Q1 = 400 g * 4.18 J/g°C * (30°C - 80°C)
Q1 = 400 g * 4.18 J/g°C * (-50°C)
Q1 = -83600 J
The negative sign indicates that heat is lost by the water as it cools down.
The nature of the final state can be determined by comparing the final temperature (30°C) to the melting point of ice (0°C). Since the final temperature is above the melting point, the final state will be a mixture of liquid water and ice. All the ice would not have melted completely; some of it would still be in its solid state.
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