A 123 g cube of ice at 0◦C is dropped into 1.9 kg of water that was originally at 81◦C.
What is the final temperature of the water after after the ice melts? The latent heat of fusion of water is 3.33 × 105 J/kg .
Answer in units of ◦C
You have collected exactly 2800 aluminum cans for recycling, each with a mass of 13.6 g. How much energy is needed to melt them if their initial temperature is 26.2◦C? Assume the specific heat, the latent heat and the melting point of aluminum are 899 J/kg ·◦ C,
3.97 × 105 J/kg and 660.4 ◦C, respectively. Answer in units of J
Ice:
Q= λm + cm(t-0) = λm + cmt
Water
Q=cm₁(81-t)
λ=3.33•10^5 J/kg .
m=0.123 kg
m₁ = 1.9 kg
c = 4186 J/kg •◦ C
λm + cmt= c m₁(81-t)
Solve for ‚t’
To find the final temperature of the water after the ice melts, we need to consider the heat transfer that occurs during the phase change and the heat transfer that occurs afterward to bring both substances to the same final temperature.
First, let's calculate the heat transfer during the phase change of the ice. The heat transfer formula is given by:
Q = m x L
Where:
Q is the heat transfer
m is the mass of the substance
L is the specific latent heat
Since the cube of ice has a mass of 123 grams (0.123 kg) and the specific latent heat of fusion of water is given as 3.33 × 10^5 J/kg, we can calculate the heat transfer during the phase change of the ice:
Q = 0.123 kg x 3.33 × 10^5 J/kg
Q = 40959 J
Next, we need to consider the heat transfer that occurs afterward to bring both the water and the melted ice to the same final temperature.
We can use the heat transfer formula again:
Q = m x c x ΔT
Where:
Q is the heat transfer
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
The mass of the water is given as 1.9 kg, and the initial temperature of the water is 81°C. We need to find the final temperature, so let's express it as T.
Qwater = 1.9 kg x c x (T - 81)
The heat transfer for the melted ice will be:
Qice = 0.123 kg x c x (T - 0)
Since the heat transfer during the phase change (Q) is the same for both the melted ice and the water, we can equate these two equations:
Qwater = Qice
1.9 kg x c x (T - 81) = 0.123 kg x c x T
Simplifying the equation, we get:
1.9T - 154.9c = 0.123T
Bringing the T terms to one side and the c terms to the other side, we get:
1.9T - 0.123T = 154.9c
1.777T = 154.9c
T = 154.9c / 1.777
Now, we can substitute the given values into the equation to find the final temperature in Celsius:
T = 154.9 / 1.777
T ≈ 87.17°C
Therefore, the final temperature of the water after the ice melts is approximately 87.17°C.