What is the final temperature when 2.50 x 10^5 J are added to 0.950 kg of ice at 0.0 degrees C? The specific heat capacity of water is 4186 J/(kg·degrees C) and the latent heat of fusion for ice is 3.35×10^3 J/kg.
To determine the final temperature, we need to consider the heat required to raise the temperature of the ice from 0.0 degrees C to its melting point (0 degrees C), and then the heat required to melt the ice completely.
Step 1: Heat required to raise temperature to 0 degrees C
The specific heat capacity of ice is 2100 J/(kg·degrees C), given that ice is at 0.0 degrees C and we need to raise its temperature to 0 degrees C. The formula for heat is:
Q = mcΔT
where Q is the heat required, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
Q1 = (0.950 kg) * (2100 J/(kg·degrees C)) * (0 degrees C - 0.0 degrees C)
Q1 = 0 J
Step 2: Heat required to melt ice
The latent heat of fusion for ice is 3.35×10^3 J/kg. Since the entire mass of ice is going to melt, the heat required is:
Q2 = (0.950 kg) * (3.35×10^3 J/kg)
Q2 = 3.1825×10^3 J
Step 3: Heat required to raise temperature of water to final temperature
Once the ice has melted, we need to calculate the heat required to raise the temperature of the resulting water from 0 degrees C to the final temperature.
Q3 = (0.950 kg) * (4186 J/(kg·degrees C)) * (final temperature - 0 degrees C)
Q3 = 0.950 kg * 4186 J/(kg·degrees C) * final temperature
Step 4: Total heat added
Now, we can calculate the total heat added to the system, which is the sum of the three steps above.
Total heat added = Q1 + Q2 + Q3
Total heat added = 0 J + 3.1825×10^3 J + 0.950 kg * 4186 J/(kg·degrees C) * final temperature
Total heat added = 3.1825×10^3 J + 0.950 kg * 4186 J/(kg·degrees C) * final temperature
Step 5: Solve for the final temperature
Since we know the total heat added is 2.50 x 10^5 J, we can set up an equation and solve for the final temperature.
2.50 x 10^5 J = 3.1825×10^3 J + 0.950 kg * 4186 J/(kg·degrees C) * final temperature
Rearranging the equation:
2.50 x 10^5 J - 3.1825×10^3 J = 0.950 kg * 4186 J/(kg·degrees C) * final temperature
2.46675×10^5 J = 0.950 kg * 4186 J/(kg·degrees C) * final temperature
Dividing both sides of the equation by (0.950 kg * 4186 J/(kg·degrees C)):
final temperature = 2.46675×10^5 J / (0.950 kg * 4186 J/(kg·degrees C))
final temperature ≈ 56.85 degrees C
Therefore, the final temperature is approximately 56.85 degrees C.
To find the final temperature when energy is added to ice, we need to consider the energy required to raise the ice temperature from 0.0 degrees C to its melting point, and then the energy required to change the ice into water at its melting point.
First, we calculate the energy required to raise the temperature of ice using the specific heat capacity formula:
Q1 = m * c * ΔT
Where:
Q1 = energy required to raise the temperature
m = mass of the ice = 0.950 kg
c = specific heat capacity of water = 4186 J/(kg·degrees C)
ΔT = change in temperature = final temperature - initial temperature
ΔT = T - (0.0 degrees C)
Next, we calculate the energy required to change the ice into water using the latent heat of fusion formula:
Q2 = m * L
Where:
Q2 = energy required to change the state of the ice into water
L = latent heat of fusion for ice = 3.35×10^3 J/kg
Finally, we sum up both energy values to determine the total energy added to the system:
Total energy = Q1 + Q2
To find the final temperature, we rearrange the equation and solve for T:
T = (Total energy / (m * c)) + (0.0 degrees C)
Note: Be sure to convert the given energy from 2.50 x 10^5 J to J before starting the calculations.
Now you can use the above formulas and calculations to find the final temperature.
Please use steps!! Thanks
How much heat is required to melt the ice at 0C? That's
0.950 kg x 3.35E3 J/kg = ?
Subtract that from 2.50E5 J to determine the excess heat available to heat the 0.950 kg water at 0C left. Then
q(what's left) = mass x specific heat liquid H2O x (Tfinal-Tinitial) and solve for Tfinal.
The answer is approx 60 C.