A 0.0475 kg ice cube at −30.0°C is placed in 0.577 kg of 35.0°C water in a very well insulated container. The latent heat of fusion of water is 334 kJ/kg, the specific heat of ice is 2092 J/(kg · K), and the specific heat of water is 4184 J/(kg · K). The system comes to equilibrium after all of the ice has melted. What is the final temperature of the system?
HINTS: This problem must be broken into pieces. Think about it chronologically: The ice can't melt until it warms up to its melting point. The ice will then melt, but will still be at zero degrees. You will need to then consider the melted ice warming to the equilibrium temperature, and the original water cooling to the equilibrium temperature.
Nuts to the hint. The sum of the heats gained in the system is zero (some gain, some lose)
Heatwarmingice+Heatfusion+heatwarming melted ice+heat warming orig water=0
let mi be the massice, mw mass of the orginal water.
mi*Ci*(0+30+ mi*Li+ mi(cw(tf-30)+mw*cw*(Tf-35)=0
solve for Tf
To solve this problem, let's break it down into smaller steps and consider the energy changes at each stage.
Step 1: Ice warming up to its melting point
Since the ice needs to warm up to its melting point of 0°C before it can start melting, we can calculate the energy change using the specific heat of ice formula:
Q1 = m × c_ice × ΔT
where:
- Q1 is the energy change
- m is the mass of the ice (0.0475 kg)
- c_ice is the specific heat of ice (2092 J/(kg · K))
- ΔT is the change in temperature
The ice is initially at -30.0°C and needs to warm up to 0°C, so ΔT1 = 0 - (-30) = 30°C.
Now substitute the values into the equation:
Q1 = 0.0475 kg × 2092 J/(kg · K) × 30°C
Step 2: Ice melting at 0°C
Once the ice reaches its melting point, it will start melting. The energy required for the phase change (latent heat of fusion) can be calculated as:
Q2 = m × L_fusion
where:
- Q2 is the energy change
- m is the mass of the ice (0.0475 kg)
- L_fusion is the latent heat of fusion of water (334 kJ/kg = 334,000 J/kg)
Step 3: Melted ice warming up to the equilibrium temperature
After all the ice has melted, the resulting water will still be at 0°C. We need to calculate the energy change for the water to reach the final equilibrium temperature.
Q3 = m × c_water × ΔT3
where:
- Q3 is the energy change
- m is the mass of the melted ice (0.0475 kg)
- c_water is the specific heat of water (4184 J/(kg · K))
- ΔT3 is the change in temperature from 0°C to the final temperature of the system
Step 4: Original water cooling down to the equilibrium temperature
The original water is initially at 35.0°C and needs to cool down to the final temperature. We can calculate the energy change using the specific heat of water formula:
Q4 = m × c_water × ΔT4
where:
- Q4 is the energy change
- m is the mass of the original water (0.577 kg)
- c_water is the specific heat of water (4184 J/(kg · K))
- ΔT4 is the change in temperature from 35.0°C to the final temperature of the system
Step 5: Equating the energy changes
Since the system is well insulated, the total energy change (Q_total) must be zero, as no heat is exchanged with the surroundings. Therefore, we can write:
Q_total = Q1 + Q2 + Q3 + Q4 = 0
Solving this equation will give us the final temperature of the system (ΔT).
Note: Make sure to convert all units to the same system (either all SI or all metric) before plugging in the values and calculating the energy changes.