If 200 g of ice at 0°C is added to 100 g of water at 80°C, what will be the final temperature of the mixture? (Latent heat of fusion of ice = 334 J/g)
Practice Questions
1 question
Q1
If 200 g of ice at 0°C is added to 100 g of water at 80°C, what will be the final temperature of the mixture? (Latent heat of fusion of ice = 334 J/g)
0°C
20°C
40°C
60°C
Heat lost by water = Heat gained by ice. 100g * 80°C = 200g * 334 J/g + 200g * (Tf - 0°C). Solving gives Tf = 20°C.
Questions & Step-by-step Solutions
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Q
Q: If 200 g of ice at 0°C is added to 100 g of water at 80°C, what will be the final temperature of the mixture? (Latent heat of fusion of ice = 334 J/g)
Solution: Heat lost by water = Heat gained by ice. 100g * 80°C = 200g * 334 J/g + 200g * (Tf - 0°C). Solving gives Tf = 20°C.
Steps: 7
Step 1: Identify the initial conditions. We have 200 g of ice at 0°C and 100 g of water at 80°C.
Step 2: Understand that when ice is added to water, the ice will absorb heat to melt and then warm up, while the water will lose heat.
Step 3: Write the equation for heat lost by the water. The water loses heat as it cools down from 80°C to the final temperature (Tf). The heat lost by the water can be calculated as: 100 g * (80°C - Tf).
Step 4: Write the equation for heat gained by the ice. The ice first melts and then warms up to the final temperature. The heat gained by the ice can be calculated as: (200 g * 334 J/g) + (200 g * (Tf - 0°C)).
Step 5: Set the heat lost by the water equal to the heat gained by the ice: 100 g * (80°C - Tf) = 200 g * 334 J/g + 200 g * (Tf - 0°C).
Step 6: Simplify the equation and solve for Tf. This will give you the final temperature of the mixture.
Step 7: After solving the equation, you find that Tf = 20°C.
