Let's break down your questions step by step, focusing on the principles of thermal physics, specifically heat transfer and phase changes. We'll tackle each question separately, ensuring clarity and understanding.
Question 1: Ice and Water Mixture
In this scenario, we have 140 g of ice at -15°C and 200 g of water at 40°C. The goal is to find the final temperature of the system when they reach thermal equilibrium, assuming no heat is lost to the surroundings.
Step 1: Calculate the heat required to warm the ice to 0°C
First, we need to warm the ice from -15°C to 0°C. The specific heat capacity of ice is approximately 2.09 J/g°C. The formula to calculate the heat (Q) required is:
Q = m × c × ΔT
- m = mass of ice = 140 g
- c = specific heat capacity of ice = 2.09 J/g°C
- ΔT = change in temperature = 0 - (-15) = 15°C
Now, substituting the values:
Q_ice = 140 g × 2.09 J/g°C × 15°C = 4395 J
Step 2: Calculate the heat released by the water as it cools to 0°C
Next, we need to calculate how much heat the water will lose when it cools from 40°C to 0°C. The specific heat capacity of water is about 4.18 J/g°C.
Using the same formula:
- m = mass of water = 200 g
- c = specific heat capacity of water = 4.18 J/g°C
- ΔT = change in temperature = 0 - 40 = -40°C
Now, substituting the values:
Q_water = 200 g × 4.18 J/g°C × (-40°C) = -33440 J
Step 3: Determine the heat balance
For the system to reach equilibrium, the heat gained by the ice must equal the heat lost by the water:
Q_ice + Q_water = 0
Substituting the values we calculated:
4395 J + (-33440 J) = 0
This means that the water can provide enough heat to warm the ice to 0°C, and there will still be heat left over. The remaining heat will be used to melt the ice.
Step 4: Calculate the heat required to melt the ice
The heat required to melt the ice at 0°C is given by:
Q_melt = m × L_f
- m = mass of ice = 140 g
- L_f = latent heat of fusion for ice = 334 J/g
Substituting the values:
Q_melt = 140 g × 334 J/g = 46760 J
Step 5: Final temperature and masses
Now, we need to see if the water can provide enough heat to melt all the ice. The total heat available from the water is:
Q_available = 33440 J
Since 33440 J is less than 46760 J, not all the ice will melt. We can find out how much ice melts by calculating:
melted_ice = Q_available / L_f
melted_ice = 33440 J / 334 J/g = 100 g
This means 100 g of ice melts, leaving us with:
- mass of ice remaining = 140 g - 100 g = 40 g
- mass of water = 200 g + 100 g = 300 g
The final temperature of the system will be 0°C, as the remaining ice and water will coexist at this temperature.
Question 2: Melting Ice with Water
In this case, we have equal masses of ice and water in contact, and we want to find the temperature of the water needed to completely melt the ice. Let's assume we have 100 g of ice at 0°C and 100 g of water at an unknown temperature (T).
Step 1: Heat required to melt the ice
Using the latent heat of fusion:
Q_melt = m × L_f
- m = mass of ice = 100 g
- L_f = 334 J/g
Substituting the values:
Q_melt = 100 g × 334 J/g = 33400 J
Step 2: Heat released by the water
The heat lost by the water as it cools down to 0°C is:
Q_water = m × c × ΔT
- m = mass of water = 100 g
- c = specific heat capacity of water = 4.18 J/g°C
- ΔT = change in temperature = T - 0
So, we have:
Q_water = 100 g × 4.18 J/g°C × (T - 0) = 418T J
Step 3: Setting up the heat balance
For the ice to completely melt, the heat lost by the water must equal the heat gained by the ice:
Q_water = Q_melt
Substituting the values we calculated:
