A 50 gram sample of liquid water at 25 degrees C is mixed with 29 grams of water at 45 degrees C. The final temperature of the water is...?
32. C
To find the final temperature of the water, we can use the principle of heat transfer, which states that the heat lost by the hotter object is equal to the heat gained by the colder object.
First, let's calculate the heat lost by the 45 degrees Celsius water:
Q_lost = m × c × ΔT_lost
Where:
m = mass of the water
c = specific heat capacity of water
ΔT_lost = change in temperature of the water
Since the mass of the water is 29 grams and the specific heat capacity of water is approximately 4.18 J/g°C, we can plug in the values:
Q_lost = 29 g × 4.18 J/g°C × (45°C - T_final)
Now, let's calculate the heat gained by the 25 degrees Celsius water:
Q_gained = m × c × ΔT_gained
Where:
m = mass of the water
c = specific heat capacity of water
ΔT_gained = change in temperature of the water
Since the mass of the water is 50 grams and the specific heat capacity of water is approximately 4.18 J/g°C, we can plug in the values:
Q_gained = 50 g × 4.18 J/g°C × (T_final - 25°C)
According to the principle of heat transfer, Q_lost = Q_gained. Therefore, we can set the two equations equal to each other and solve for the final temperature, T_final:
29 g × 4.18 J/g°C × (45°C - T_final) = 50 g × 4.18 J/g°C × (T_final - 25°C)
Now let's solve the equation:
29 × 4.18 × (45 - T_final) = 50 × 4.18 × (T_final - 25)
Simplifying further:
1211 - 29 × 4.18 × T_final = 2090 × T_final - 2090 × 25
Combining like terms:
348.52 × T_final = 52550 - 1211
348.52 × T_final = 51339
T_final = 51339 / 348.52
T_final ≈ 147.25°C
Therefore, the final temperature of the water is approximately 147.25 degrees Celsius.
To find the final temperature of the water after mixing, you can use the principle of conservation of energy, specifically the equation for heat transfer:
Q = mcΔT
where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat transferred from the hot water:
Q1 = m1c1ΔT1
where m1 = 29g (mass of hot water), c1 is the specific heat capacity of water, and ΔT1 = Tf - 45°C (change in temperature from the initial hot water temperature to the final temperature Tf).
Second, let's calculate the heat transferred from the cold water:
Q2 = m2c2ΔT2
where m2 = 50g (mass of cold water), c2 is the specific heat capacity of water, and ΔT2 = Tf - 25°C (change in temperature from the initial cold water temperature to the final temperature Tf).
Since heat is conserved when the two water samples mix, the heat transferred from the hot water (Q1) is equal to the heat transferred to the cold water (Q2):
Q1 = Q2
m1c1ΔT1 = m2c2ΔT2
Plugging in the given values, we have:
29g * c1 * (Tf - 45°C) = 50g * c2 * (Tf - 25°C)
Now, we can simplify the equation and solve for Tf, the final temperature:
29c1Tf - 29c1(45°C) = 50c2Tf - 50c2(25°C)
29c1Tf - 1305c1 = 50c2Tf - 1250c2
(29c1Tf - 50c2Tf) = (1305c1 - 1250c2)
Tf(29c1 - 50c2) = 1305c1 - 1250c2
Tf = (1305c1 - 1250c2) / (29c1 - 50c2)
Now, to find the final temperature, we need to know the specific heat capacities of water, which are:
c1 = 4.18 J/g°C (for hot water)
c2 = 4.18 J/g°C (for cold water)
Plugging in these values, we get:
Tf = (1305 * 4.18 - 1250 * 4.18) / (29 * 4.18 - 50 * 4.18)
Calculating the equation, we find:
Tf = 44.46°C
Therefore, the final temperature of the water after mixing is 44.46°C.
heat gained by cool H2O + heat lost by warm H2O = 0
[mass cool H2O x specific heat H2O x (Tfinal-Tintial)] + [mass warm water x specific heat H2O x (Tfinal-Tintial)] = 0
Substitute and solve for Tf