a student mixes 88.6g of water at 74.3 degree Celsius with 57.9g of water at 24.8 degree Celsius in an insulated flask.What is the final temperature of the combined water.
54.7 degrees C
To find the final temperature of the combined water, we can use the principle of conservation of energy. The heat gained by the colder water is equal to the heat lost by the hotter water.
The heat gained by the colder water (Q1) can be calculated using the formula:
Q1 = mass1 * specific heat capacity * change in temperature1
where,
mass1 = mass of colder water = 57.9g
specific heat capacity = specific heat capacity of water = 4.18 J/g°C (approximate value)
change in temperature1 = final temperature - initial temperature of colder water = final temperature - 24.8°C
Similarly, the heat lost by the hotter water (Q2) can be calculated using the formula:
Q2 = mass2 * specific heat capacity * change in temperature2
where,
mass2 = mass of hotter water = 88.6g
specific heat capacity = specific heat capacity of water = 4.18 J/g°C (approximate value)
change in temperature2 = initial temperature of hotter water - final temperature
Since the two masses of water are equal, the amount of heat gained by the colder water (Q1) is equal to the amount of heat lost by the hotter water (Q2):
Q1 = Q2
Using the given values, we can set up the equation and solve for the final temperature:
mass1 * specific heat capacity * change in temperature1 = mass2 * specific heat capacity * change in temperature2
(57.9g) * (4.18 J/g°C) * (final temperature - 24.8°C) = (88.6g) * (4.18 J/g°C) * (74.3°C - final temperature)
Simplifying the equation:
(57.9g) * (final temperature - 24.8°C) = (88.6g) * (74.3°C - final temperature)
Solving for the final temperature:
57.9g * final temperature - (57.9g * 24.8°C) = 88.6g * 74.3°C - 88.6g * final temperature
(57.9g + 88.6g) * final temperature = (88.6g * 74.3°C) + (57.9g * 24.8°C)
146.5g * final temperature = 6567.98g°C
final temperature = 6567.98g°C / 146.5g
final temperature ≈ 44.86°C
Therefore, the final temperature of the combined water is approximately 44.86°C.
To find the final temperature of the combined water, we can use the principle of conservation of energy. The total energy before and after the mixing of the water remains the same.
To calculate the final temperature, we need to use the equation for heat transfer:
Q = mcΔT
Where:
Q = Heat transferred (in calories or joules)
m = Mass of the substance (in grams or kilograms)
c = Specific heat capacity of the substance (in calories/gram°C or joules/kg°C)
ΔT = Change in temperature (in °C)
First, let's find the heat transfer for each sample of water using the equation:
Q1 = m1c1ΔT1 (for the initial 88.6g of water)
Q2 = m2c2ΔT2 (for the initial 57.9g of water)
The specific heat capacity of water is approximately 1 calorie/gram°C or 4.18 joules/gram°C.
For the first sample of water:
Q1 = (88.6g)(1 cal/g°C)(Tf - 74.3°C)
For the second sample of water:
Q2 = (57.9g)(1 cal/g°C)(Tf - 24.8°C)
Since the flask is insulated, the heat transferred from one sample of water to the other is equal. So we can set Q1 equal to Q2:
(88.6g)(1 cal/g°C)(Tf - 74.3°C) = (57.9g)(1 cal/g°C)(Tf - 24.8°C)
Now we can solve for Tf, the final temperature of the combined water. Let's simplify the equation:
88.6(Tf - 74.3) = 57.9(Tf - 24.8)
88.6Tf - 6569.98 = 57.9Tf - 1431.92
Combining like terms:
88.6Tf - 57.9Tf = 6569.98 - 1431.92
30.7Tf = 5138.06
Dividing both sides by 30.7:
Tf = 167.46 °C (approximately)
Therefore, the final temperature of the combined water is approximately 167.46 degrees Celsius.
heat lost by warm water + heat gained by cool water = 0
[mass warm water x specific heat water x (Tfinal-Tinitial)] + [mass cool water x specific heat water x T(final-Tinitial)] = 0
Solve for Tfinal.