in a calorimeter can(water equivalent=40g)are 200g of water and 50g of ice,all at 0 degreea celcius.into these poured 30g of water at 90degrees celcuis.what will be the final temperature of the system?specific capacity of water is 4186j/kg per degrees celcius,of ice is 2310j/kg per degrees celcuis,and latent heat of fusion is 334000j/kg
The heat-energy needed to produce any temperature change in the liquid is: DeltaH = mass x heat-capacity x DeltaT
where DeltaT is the change in temperature.
The heat-energy needed to melt the ice is: Hmelting = mass x heat-of-fusion
Lets assume the final temperature is 'T'. Then...
The energy change in the 30g of hot water is:
DeltaH(hotwater) = 30g x (4186j/kg-C*1kg/1000g) x (T-90deg)
The energy change in the 200g of cold water is:
DeltaH(coldwater) = 200g x (4186j/kg-C*1kg/1000g) x (T-0deg)
The energy change in the can is:
DeltaH(can) = 40g x (4186j/kg-C*1kg/1000g) x (T-0deg)
Note: The can should start at the same temperature as the water and ice it contains, so initial temperature=0deg C.
Finally, the energy change in the ice is (assuming the ice melts completely):
DeltaH(ice) = 50g x (334000j/kg*1kg/1000g) + 50g x (4186j/kg-C*1kg/1000g) x (T-0deg)
Because energy is conserved all these changes in energy in the individual components of the system should sum to zero for the whole system:
DeltaH(hotwater)+DeltaH(coldwater)+DeltaH(can)+DeltaH(ice) = 0
So we get:
30g x (4186j/kg-C*1kg/1000g) x (T-90deg)
+
200g x (4186j/kg-C*1kg/1000g) x (T-0deg)
+
40g x (4186j/kg-C*1kg/1000g) x (T-0deg)
+
50g x (334000j/kg*1kg/1000g)
+ 50g x (4186j/kg-C*1kg/1000g) x (T-0deg)
= 0
Now we just have to solve for 'T'.
Simplifying, we get:
30g x 4.186j/g-C x (T-90)
+(200+40+50)g x 4.186j/g-C x (T-0)
+50g x (334j/g) = 0
OR
(30+200+40+50)g x 4.186j/g-C x T
+ 30g x 4.186j/g-C x (-90)
+ 50g x 334j/g = 0
OR
320g x 4.186j/g-C x T
+ -11302.2j
+ 16700j = 0
OR
1339.52j/C x T = -5397.8j
OR
T = -5397.8j/1339.52j/C = -4.029dec C
BUT! This answer is wrong. If we add hot water to cold water and ice the result can never be colder than the starting point (0dec C). This means one of our assumptions was wrong.
We only made two assumptions:
1) the can is the same temperature as the water and ice at the start (0deg)
2) the ice melts completely
The 2nd assumption must be incorrect, so we need to change our calculations accordingly:
If the ice does not melt completely, then there will still be ice left in the system. If ice is left then the final temperature must be 0deg celcius because ice and water at equilibrium must be at 0deg celcius (the melting point, by definition).
A related question then is how much ice melted? This can be solved in a similar way:
DeltaH(hotwater)= 30g x (4186j/kg-C*1kg/1000g) x (0-90deg)
DeltaH(hotwater)= -11302.2j
DeltaH(ice) = mass? x (334000j/kg*1kg/1000g)
Energy balance tells us:
DeltaH(hotwater)+DeltaH(ice) = 0
OR
-11302.2j + mass? x 334000j/kg*1kg/1000g = 0
OR mass? x 334j/g = 11302.2j
OR mass? = 11302.2j / 334j/g
OR mass? = 33.84g
So:
33.84g of ice melted after adding 30g of hot water to the system, but the system is still at 0deg celcius.
the sum of the heats gained is zero (some will be lost).
masscan*cw*(0-Tf)+200*Cw*(0-Tf)+50*Hfice+ 50*cw*(0-tf)+30*cw*(Tf-90)=0
solve for Tf.
To find the final temperature of the system, we need to consider the heat transfer between the water, ice, and the added hot water using the principle of conservation of energy.
The total heat gained by the water and ice = heat lost by the hot water.
First, let's calculate the heat gained by the water and ice:
For water:
Heat gained by water = mass of water x specific heat capacity of water x change in temperature
= 200g x 4186 J/kg°C x (final temperature - 0°C)
For ice:
Heat gained by ice = mass of ice x specific heat capacity of ice x change in temperature
= 50g x 2310 J/kg°C x (final temperature - 0°C)
Now, let's calculate the heat lost by the hot water:
Heat lost by hot water = mass of hot water x specific heat capacity of water x change in temperature
= 30g x 4186 J/kg°C x (90°C - final temperature)
Since the ice will be melting, it will also absorb latent heat, which can be calculated as:
Latent heat absorbed by ice = mass of ice x latent heat of fusion
= 50g x 334000 J/kg
Using the principle of conservation of energy:
Heat gained by water + Heat gained by ice + Latent heat absorbed by ice = Heat lost by hot water
(200g x 4186 J/kg°C x (final temperature - 0°C)) + (50g x 2310 J/kg°C x (final temperature - 0°C)) + (50g x 334000 J/kg) = (30g x 4186 J/kg°C x (90°C - final temperature))
Now, solve the equation to find the final temperature of the system.
To find the final temperature of the system, we need to calculate the amount of heat gained or lost by each component and use the principle of conservation of energy.
First, let's calculate the heat gained or lost by the ice during the phase change from solid to liquid (melting).
The heat gained during the phase change (Q1) can be calculated using the equation:
Q1 = mass of ice × latent heat of fusion
Given that the mass of ice (m1) is 50g and the latent heat of fusion (L1) is 334,000 J/kg, we can calculate Q1 as follows:
Q1 = 0.05kg × 334,000 J/kg = 16,700 J
Next, let's calculate the heat gained or lost by the ice as it warms up from the initial temperature of 0 degrees Celsius (T1) to the final temperature of the system (Tf).
The heat gained during this warming phase (Q2) can be calculated using the equation:
Q2 = mass of ice × specific heat capacity of ice × change in temperature
Given that the specific heat capacity of ice (C1) is 2,310 J/(kg·°C), and the change in temperature (ΔT) is Tf - T1, we can calculate Q2 as follows:
Q2 = 0.05kg × 2,310 J/(kg·°C) × Tf
Now, let's calculate the heat gained by the water as it warms up from the initial temperature of 0 degrees Celsius (Tw) to the final temperature of the system (Tf).
The heat gained by the water (Q3) can be calculated using the equation:
Q3 = mass of water × specific heat capacity of water × change in temperature
Given that the mass of water (m2) is 200g, the specific heat capacity of water (C2) is 4,186 J/(kg·°C), and the change in temperature (ΔT) is Tf - Tw, we can calculate Q3 as follows:
Q3 = 0.2kg × 4,186 J/(kg·°C) × (Tf - 0)
Now, let's calculate the heat lost by the 30g of water at 90 degrees Celsius (T4) as it cools down to the final temperature of the system (Tf).
The heat lost by this water (Q4) can be calculated using the equation:
Q4 = mass of water × specific heat capacity of water × change in temperature
Given that the mass of water (m3) is 30g, the specific heat capacity of water (C2) is 4,186 J/(kg·°C), and the change in temperature (ΔT) is 90 - Tf, we can calculate Q4 as follows:
Q4 = 0.03kg × 4,186 J/(kg·°C) × (90 - Tf)
The total heat gained by the system is equal to the total heat lost by the 30g of water at 90 degrees Celsius. Therefore,
Q1 + Q2 + Q3 = Q4
Substituting the values we calculated earlier, we get:
16,700 J + (0.05kg × 2,310 J/(kg·°C) × Tf) + (0.2kg × 4,186 J/(kg·°C) × Tf) = 0.03kg × 4,186 J/(kg·°C) × (90 - Tf)
Now, we can solve this equation to find the final temperature of the system (Tf). It involves some algebraic manipulation, which can be solved using numerical methods or software.
Note: It's important to mention that the water equivalent of the calorimeter (40g) doesn't affect the final temperature calculation in this problem. The water equivalent is used to account for the heat absorbed by the calorimeter itself when calculating the total heat transferred in some situations. Here, we are only interested in the final temperature of the system.