When 16 g of ice at 0°C is dropped into 250 g of water at 42°C, what would be the final temperature?
This is worked almost the same as the one above. See
https://www.jiskha.com/questions/1856330/what-is-the-final-temperature-of-a-mixture-of-250-g-of-water-at-33-c-into-which-3-g-of
Just omit the part from -20 T to 0C and start with the melting of ice @ 0C.
So it will be mass(250 g) water x specific heat water x (Tfinal-Tinitial)] + [mass melted ice(16 g) x specific heat water x (Tfinal-Tinitial)] +(mass ice x heat fusion @ 0C) = 0 and solve for Tfinal.
To find the final temperature, we can use the principle of energy conservation. The heat lost by the water should be equal to the heat gained by the ice.
1. Calculate the heat lost by the water:
Qwater = mwater * cwater * ΔTwater
where,
mwater = mass of water = 250 g
cwater = specific heat capacity of water = 4.18 J/g°C
ΔTwater = change in temperature = final temperature - initial temperature = final temperature - 42°C
2. Calculate the heat gained by the ice:
Qice = mice * cice * ΔTice
where,
mice = mass of ice = 16 g
cice = specific heat capacity of ice = 2.09 J/g°C
ΔTice = change in temperature = final temperature - initial temperature = final temperature - 0°C
3. Since the heat lost by the water is equal to the heat gained by the ice, we can write:
Qwater = Qice
mwater * cwater * ΔTwater = mice * cice * ΔTice
4. Substitute the given values and solve for the final temperature:
250 g * 4.18 J/g°C * (final temperature - 42°C) = 16 g * 2.09 J/g°C * (final temperature - 0°C)
Expanding the equation:
(250 * 4.18 * final temperature) - (250 * 4.18 * 42) = (16 * 2.09 * final temperature)
Simplifying the equation:
(1045 * final temperature) - (1045 * 42) = (33.44 * final temperature)
Expanding further:
1045 * final temperature - 43790 = 33.44 * final temperature
Subtracting 33.44 * final temperature from both sides and moving the constant term to the right side, we get:
1045 * final temperature - 33.44 * final temperature = 43790
(1045 - 33.44) * final temperature = 43790
1011.56 * final temperature = 43790
Divide both sides by 1011.56 to solve for the final temperature:
final temperature = 43790 / 1011.56
Using a calculator, we find that the final temperature is approximately 43.3°C.
To find the final temperature, we can use the principle of conservation of energy.
Step 1: Calculate the heat gained by the ice.
When the ice melts, it requires a certain amount of heat energy. This is given by the formula:
Q = m × L
Where:
Q = heat gained or lost
m = mass of the substance
L = latent heat of the substance
For ice, the latent heat (L) is 334 J/g.
Given that the mass (m) of the ice is 16 g, we can calculate the heat gained:
Q_ice = 16 g × 334 J/g = 5344 J
Step 2: Calculate the heat lost by the water.
When the water cools down, it loses heat energy. This is given by the formula:
Q = m × c × ΔT
Where:
Q = heat gained or lost
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
For water, the specific heat capacity (c) is 4.18 J/g°C.
Given that the mass (m) of the water is 250 g and the initial temperature (Ti) of the water is 42°C, we can calculate the heat lost:
Q_water = 250 g × 4.18 J/g°C × (42°C - TF)
where TF is the final temperature.
Step 3: Apply the principle of conservation of energy.
According to the principle of conservation of energy, the heat gained by the ice must be equal to the heat lost by the water:
Q_ice = Q_water
5344 J = 250 g × 4.18 J/g°C × (42°C - TF)
Step 4: Solve for the final temperature (TF).
Rearranging the equation from Step 3:
42°C - TF = 5344 J / (250 g × 4.18 J/g°C)
42°C - TF = 5.08°C
Now, we can solve for TF:
TF = 42°C - 5.08°C = 36.92°C
Therefore, the final temperature after the ice is dropped into the water would be approximately 36.92°C.