Thermodynamics Practice Problems Answer The Following Questi ✓ Solved
Thermodynamics Practice Problems Answer the following questions
1) An electric kettle containing 1.5 kg of water is heated from 15 oC to 100 oC. The specific heat of water is 4180 , calculate the amount of energy required to raise the temperature of the water. If it takes 4 minutes to heat the kettle what is the power required?
2) Ice is placed in a water pitcher which is at room temperature (23 oC). The specific heat of water is 4180 and the specific heat of ice is 2220 . If there are 200 g of ice and 1.5 kg of water, what is the final temperature of the water just after all the ice melts (when the temperature has reached equilibrium).
3) A particular type of industrial ice maker can produce a 1kg block of ice. How much power is required if it takes 45 minutes for the ice block to freeze? Assume the water starts at room temperature (23 oC) and has a specific heat of 4180 . (Hint think about solving for Q first. What is the relationship between Q and P?)
Paper For Above Instructions
Thermodynamics is a branch of physics that deals with heat, work, and energy. It is essential in many applications, including environmental science, engineering, and even everyday cooking. In this paper, we will solve three thermodynamic problems using the fundamental formulas pertinent to energy transfer, heating, and phase change.
Problem 1: Heating Water in an Electric Kettle
The first problem involves heating 1.5 kg of water from an initial temperature of 15 °C to a final temperature of 100 °C. To calculate the amount of energy (Q) required for this process, we can use the formula:
Q = mcΔT
Where:
- m = mass of the water (1.5 kg)
- c = specific heat capacity of water (4180 J/kg°C)
- ΔT = change in temperature (final temperature - initial temperature) = (100 °C - 15 °C) = 85 °C
Now we can plug in the values:
Q = 1.5 kg 4180 J/kg°C 85 °C
Q = 1.5 4180 85 = 533,250 J
This means that to heat 1.5 kg of water from 15 °C to 100 °C, approximately 533,250 Joules of energy is required.
Next, we will determine the power required (P) to achieve this heating in 4 minutes (240 seconds). Power can be calculated using the relationship:
P = Q/t
Here, t is the time in seconds. Substituting the known values:
P = 533,250 J / 240 s = 2,218.75 W
Thus, the power required to heat the kettle is about 2,218.75 Watts.
Problem 2: Melting Ice in Water
The second problem involves a scenario where 200 g of ice at 0 °C is placed into 1.5 kg of water at room temperature (23 °C). To determine the final equilibrium temperature of the water once all the ice has melted, we need to consider both the specific heat of water and the latent heat for melting ice.
First, we convert the mass of ice to kilograms for consistency:
Mass of ice, m_ice = 0.2 kg
Mass of water, m_water = 1.5 kg
The specific heat of water (c_water) is 4180 J/kg°C, and the specific heat of ice (c_ice) is 2220 J/kg°C. The latent heat of fusion (L_f) for ice is approximately 334,000 J/kg.
When the ice melts, it absorbs heat (Q_ice) equal to:
Q_ice = m_ice L_f = 0.2 kg 334,000 J/kg = 66,800 J
Now, let us determine how much energy the water loses as it cools down to the final temperature (T_final).
Energy lost by water (Q_water) is given by:
Q_water = m_water c_water (T_initial - T_final) = 1.5 kg 4180 J/kg°C (23 °C - T_final)
Equating the heat gained by ice to the heat lost by water gives:
Q_ice = Q_water
66,800 J = 1.5 kg 4180 J/kg°C (23 °C - T_final)
This can be solved to find T_final:
66,800 J = 6270 * (23 - T_final)
T_final = 23 - (66,800 / 6270) = 23 - 10.67 ≈ 12.33 °C
The final temperature of the system after all the ice has melted is approximately 12.33 °C.
Problem 3: Power Required for Ice Maker
The third problem involves calculating the power needed for an industrial ice maker that produces a 1 kg block of ice in 45 minutes. To solve this, we need to first determine the amount of energy required to freeze 1 kg of water at room temperature (23 °C) to ice at 0 °C.
The energy needed (Q) to lower the temperature of the water to 0 °C is:
Q = mcΔT
For this case:
- m = mass of the water (1 kg)
- c = specific heat of water (4180 J/kg°C)
- ΔT = change in temperature = (0 °C - 23 °C) = -23 °C
Substituting values, we get:
Q = 1 kg 4180 J/kg°C (-23 °C) = -96,140 J
This is the energy required to cool the water to 0 °C. However, in addition, we must also account for the latent heat required to freeze the water:
Q_freeze = m L_f = 1 kg 334,000 J/kg = 334,000 J
The total energy required (Q_total) is:
Q_total = |Q| + Q_freeze = 96,140 J + 334,000 J = 430,140 J
Now, let's calculate the power (P) required using the total energy over the time:
Time taken is 45 minutes = 2700 seconds. Now, using the formula:
P = Q_total / t = 430,140 J / 2700 s ≈ 159.3 W
Thus, the power required for the ice maker to produce a 1 kg block of ice is roughly 159.3 Watts.
Conclusion
In summary, we examined three thermodynamic problems involving the heating of water, the melting of ice, and the freezing of water into ice. By applying the specific heat equations and understanding the relationships between energy, mass, temperature change, and time, we found that energy and power calculations are crucial in many real-world applications such as cooking and industrial processes.
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