If 200 g of ice at 0°C is added to 100 g of water at 80°C, what will be the fina
Practice Questions
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
Questions & Step-by-Step Solutions
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)
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.
Heat Transfer – The principle of conservation of energy where heat lost by the warmer substance equals heat gained by the cooler substance.
Latent Heat – The amount of heat required to change the state of a substance without changing its temperature, specifically the heat of fusion in this case.
Thermal Equilibrium – The state reached when two substances at different temperatures come into contact and exchange heat until they reach the same temperature.
