A 46 g ice cube at 0◦C is placed in 988 g of water at 90◦C. What is the final temperature of the mixture? The specific heat of water is 4186 J/kg · ◦C and its latent heat of fusion is 3.33 × 105 J/kg .
Answer in units of ◦C
m(ice)=m₁=0.046 kg
L=3.33•10⁵J/kg
m(water) =m₂=0.988 kg
c=4186 J/kg•℃
m₁L+m₁c(t-0) =m₂c(90-t)
Solve for t
To find the final temperature of the mixture, we can use the principle of conservation of energy. The heat lost by the water at 90◦C will be equal to the heat gained by the ice cube at 0◦C.
First, let's calculate the heat lost by the water. The heat lost can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat lost
m is the mass of the water
c is the specific heat of water
ΔT is the change in temperature
Given:
m = 988 g = 0.988 kg (convert grams to kilograms)
c = 4186 J/kg · ◦C
ΔT = 90◦C - Tf (final temperature)
Q = 0.988 * 4186 * (90 - Tf)
Next, let's calculate the heat gained by the ice cube. The heat gained can be calculated using the formula:
Q = m * L
Where:
Q is the heat gained
m is the mass of the ice cube
L is the latent heat of fusion
Given:
m = 46 g = 0.046 kg (convert grams to kilograms)
L = 3.33 × 10^5 J/kg
Q = 0.046 * 3.33 × 10^5
Now, since the heat lost by the water is equal to the heat gained by the ice cube, we can set up an equation:
0.988 * 4186 * (90 - Tf) = 0.046 * 3.33 × 10^5
Solving this equation will give us the final temperature (Tf):
0.988 * 4186 * (90 - Tf) = 0.046 * 3.33 × 10^5
Now you can solve this equation to find the final temperature (Tf).