0.1kg of ice initially at a temperature of -10 degrees Celsius is added to a cup with .5kg of water initially at 20 degrees Celsius. The water and ice are isolated thermally from their surroundings. The specific heat of ice is 2000 J/kgoC, the specific heat of water is 4186 J/kgoC and the latent heat of fusion of water is 33.5x104 J/kg
What is the final temperature of the water?
8.33 degrees
no
To find the final temperature of the water, we need to determine the heat lost by the water and the heat gained by the ice. Since the ice is at a lower temperature than the water, heat will flow from the water to the ice until they reach the same final temperature.
Step 1: Calculate the heat lost by the water.
The heat lost by the water can be calculated using the equation:
Q = m * c * ΔT
where Q is the heat lost, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
m = 0.5 kg (mass of water)
c = 4186 J/kg°C (specific heat capacity of water)
ΔT = final temperature - initial temperature = final temperature - 20°C (since the initial temperature of water is 20°C)
Q1 = 0.5 kg * 4186 J/kg°C * (final temperature - 20°C)
Step 2: Calculate the heat gained by the ice.
The ice will absorb heat until it reaches its melting point (0°C), then it will melt while maintaining a constant temperature.
The heat gained by the ice consists of two parts:
- Heat absorbed to raise the temperature from -10°C to 0°C
- Latent heat of fusion required to melt the ice at 0°C
Heat absorbed to raise the temperature:
Using the equation Q = m * c * ΔT,
Q2 = 0.1 kg * 2000 J/kg°C * (0°C - (-10°C))
Latent heat of fusion to melt the ice:
Q3 = m * L
where L is the latent heat of fusion of water.
Given:
m = 0.1 kg (mass of ice)
L = 33.5 × 10^4 J/kg (latent heat of fusion of water)
Q3 = 0.1 kg * (33.5 × 10^4 J/kg)
Step 3: Set up the equations to solve for the final temperature.
Since energy is conserved in a closed system, the heat lost by the water should be equal to the heat gained by the ice.
Q1 = Q2 + Q3
0.5 * 4186 * (final temperature - 20) = 0.1 * 2000 * (0 - (-10)) + 0.1 * (33.5 × 10^4)
At this point, we can solve for the final temperature by rearranging the equation and solving for (final temperature - 20).
Finally, add 20°C to the value of (final temperature - 20) to find the actual final temperature of the water.