Water at 50 degrees C can be obtained by mixing together which one or more of the following combinations? Which of the others would have final temperatures higher than 50 degrees C and which lower than 50 degrees C? What will be the final temperatures for these combinations?
a. 1 kilogram of ice at 0 degrees C and 1 kilogram of steam at 100
degrees C.
b. 1 kilogram of ice at 0 degrees C and 1 kilogram of water at 100
degrees C.
c. 1 kilogram of water at 0 degrees C and 1 kilogram of steam at 100
degrees C.
d. 1 kilogram of water at 0 degrees C and 1 kilogram of water at 100
degrees C.
D ofcourse
To determine which combinations can obtain water at 50 degrees Celsius, we need to consider the principles of heat transfer and the specific heat capacities of the substances involved.
a. In this combination, we have 1 kilogram of ice at 0 degrees Celsius and 1 kilogram of steam at 100 degrees Celsius.
To calculate the final temperature, we can use the formula:
m1c1(Tf - T1) = m2c2(T2 - Tf)
Where:
m1 and m2 are the masses of the substances (in this case, 1 kilogram each)
c1 and c2 are the specific heat capacities of the substances (specific heat capacity of ice is 2.09 J/g°C and specific heat capacity of steam is 2.03 J/g°C)
Tf is the final temperature
T1 and T2 are the initial temperatures of the substances
Plugging in the values into the formula, we get:
(1 kg)(2.09 J/g°C)(Tf - 0°C) = (1 kg)(2.03 J/g°C)(100°C - Tf)
Simplifying the equation:
2.09 Tf = 203 - 2.03 Tf
4.12 Tf = 203
Tf ≈ 49.27°C
Hence, the final temperature for this combination is approximately 49.27 degrees Celsius. Therefore, it is lower than 50 degrees Celsius.
b. In this combination, we have 1 kilogram of ice at 0 degrees Celsius and 1 kilogram of water at 100 degrees Celsius.
Using the same formula as before:
(1 kg)(2.09 J/g°C)(Tf - 0°C) = (1 kg)(4.18 J/g°C)(100°C - Tf)
Simplifying the equation:
2.09 Tf = 418 - 4.18 Tf
6.27 Tf = 418
Tf ≈ 66.67°C
The final temperature for this combination is approximately 66.67 degrees Celsius, which is higher than 50 degrees Celsius.
c. In this combination, we have 1 kilogram of water at 0 degrees Celsius and 1 kilogram of steam at 100 degrees Celsius.
Using the same formula as before:
(1 kg)(4.18 J/g°C)(Tf - 0°C) = (1 kg)(2.03 J/g°C)(100°C - Tf)
Simplifying the equation:
4.18 Tf = 203 - 2.03 Tf
6.21 Tf = 203
Tf ≈ 32.69°C
The final temperature for this combination is approximately 32.69 degrees Celsius, which is lower than 50 degrees Celsius.
d. In this combination, we have 1 kilogram of water at 0 degrees Celsius and 1 kilogram of water at 100 degrees Celsius.
Using the same formula as before:
(1 kg)(4.18 J/g°C)(Tf - 0°C) = (1 kg)(4.18 J/g°C)(100°C - Tf)
Simplifying the equation:
4.18 Tf = 418 - 4.18 Tf
8.36 Tf = 418
Tf ≈ 50°C
The final temperature for this combination is approximately 50 degrees Celsius.
In summary:
a. Final temperature ≈ 49.27°C (lower than 50°C)
b. Final temperature ≈ 66.67°C (higher than 50°C)
c. Final temperature ≈ 32.69°C (lower than 50°C)
d. Final temperature ≈ 50°C (equal to 50°C)
To find the final temperature after mixing two substances, you can use the principle of energy conservation which states that the total energy before mixing is equal to the total energy after mixing.
In this case, we can use the formula:
(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0
where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Let's analyze each combination:
a. 1 kilogram of ice at 0 degrees C and 1 kilogram of steam at 100 degrees C.
For ice to reach 50 degrees C, it needs to absorb heat. The mass of ice is 1 kg, and its specific heat capacity is 2.09 J/g°C. Assuming the steam condenses completely to water, the mass of steam would be 1 kg, and its specific heat capacity is 4.18 J/g°C.
Using the formula, we get:
(1 kg * 2.09 J/g°C * ΔT1) + (1 kg * 4.18 J/g°C * (-50°C - 100°C)) = 0
Solving this equation gives us ΔT1 = 107.3°C. This means the final temperature is 107.3 degrees Celsius.
b. 1 kilogram of ice at 0 degrees C and 1 kilogram of water at 100 degrees C.
For ice to reach 50 degrees C, it needs to absorb heat. The mass of ice is 1 kg, and its specific heat capacity is 2.09 J/g°C. The mass of water is 1 kg, and its specific heat capacity is 4.18 J/g°C.
Using the formula, we get:
(1 kg * 2.09 J/g°C * ΔT1) + (1 kg * 4.18 J/g°C * ΔT2) = 0
Solving this equation gives us ΔT1 = ΔT2 = 0. This means the final temperature is 0 degrees Celsius.
c. 1 kilogram of water at 0 degrees C and 1 kilogram of steam at 100 degrees C.
For water to reach 50 degrees C, it needs to absorb heat. The mass of water is 1 kg, and its specific heat capacity is 4.18 J/g°C. Assuming the steam condenses completely to water, the mass of steam would be 1 kg, and its specific heat capacity is 4.18 J/g°C.
Using the formula, we get:
(1 kg * 4.18 J/g°C * ΔT1) + (1 kg * 4.18 J/g°C * (-50°C - 100°C)) = 0
Solving this equation gives us ΔT1 = 62.1°C. This means the final temperature is 62.1 degrees Celsius.
d. 1 kilogram of water at 0 degrees C and 1 kilogram of water at 100 degrees C.
For both waters to reach 50 degrees C, heat needs to be transferred from the hot water to the cold water. The mass of each water is 1 kg, and their specific heat capacity is 4.18 J/g°C.
Using the formula, we get:
(1 kg * 4.18 J/g°C * ΔT1) + (1 kg * 4.18 J/g°C * ΔT2) = 0
Solving this equation gives us ΔT1 = ΔT2 = -25°C. This means the final temperature is -25 degrees Celsius.
To summarize:
a. The final temperature is 107.3 degrees Celsius.
b. The final temperature is 0 degrees Celsius.
c. The final temperature is 62.1 degrees Celsius.
d. The final temperature is -25 degrees Celsius.