Find the amount of electrical energy expended in raising the temperature of 50 litres of water by 75° C.To what height could a weight of 5000 kg be raised with the same expenditure of energy ? Assume that heater has an efficiency of 90% and the lifting equipment has an efficiency of 70%.
Find the amount of electrical energy expended in raising the temperature of 50 litres of water by 75° C.To what height could a weight of 5000 kg be raised with the same expenditure of energy ? Assume that heater has an efficiency of 90% and the lifting equipment has an efficiency of 70%.
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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![**Problem Statement:**
Find the amount of electrical energy expended in raising the temperature of 50 liters of water by 75° C. To what height could a weight of 5000 kg be raised with the same expenditure of energy? Assume that the heater has an efficiency of 90% and the lifting equipment has an efficiency of 70%.
**Explanation:**
- **Electrical Energy Calculation:**
- Determine the energy required to heat the water using the formula:
\[
Q = m \cdot c \cdot \Delta T
\]
where:
- \( m \) is the mass of the water (50 kg for 50 liters, assuming the density is 1 kg/L),
- \( c \) is the specific heat capacity of water (approximately 4.18 J/g°C),
- \( \Delta T \) is the temperature change (75°C).
- **Efficiency Considerations:**
- The heater’s energy input must account for its efficiency.
- To find the actual electrical energy expended, divide the calculated energy by the heater efficiency (0.9).
- **Height Calculation:**
- Using the remaining energy and considering the lifting equipment efficiency, calculate the height to which 5000 kg can be raised.
- Use the gravitational potential energy formula:
\[
E = m \cdot g \cdot h
\]
where:
- \( E \) is the useful energy output (after applying both efficiencies),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²),
- \( h \) is the height.
This setup involves converting electrical energy into thermal energy and then using that energy to do mechanical work, taking into account the inefficiencies in each process.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F436b7c6b-491d-41a8-ad1d-aa61d361f983%2F892c1f4d-2530-4a73-a883-10d52bff753c%2Fxhpe7ve_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the amount of electrical energy expended in raising the temperature of 50 liters of water by 75° C. To what height could a weight of 5000 kg be raised with the same expenditure of energy? Assume that the heater has an efficiency of 90% and the lifting equipment has an efficiency of 70%.
**Explanation:**
- **Electrical Energy Calculation:**
- Determine the energy required to heat the water using the formula:
\[
Q = m \cdot c \cdot \Delta T
\]
where:
- \( m \) is the mass of the water (50 kg for 50 liters, assuming the density is 1 kg/L),
- \( c \) is the specific heat capacity of water (approximately 4.18 J/g°C),
- \( \Delta T \) is the temperature change (75°C).
- **Efficiency Considerations:**
- The heater’s energy input must account for its efficiency.
- To find the actual electrical energy expended, divide the calculated energy by the heater efficiency (0.9).
- **Height Calculation:**
- Using the remaining energy and considering the lifting equipment efficiency, calculate the height to which 5000 kg can be raised.
- Use the gravitational potential energy formula:
\[
E = m \cdot g \cdot h
\]
where:
- \( E \) is the useful energy output (after applying both efficiencies),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²),
- \( h \) is the height.
This setup involves converting electrical energy into thermal energy and then using that energy to do mechanical work, taking into account the inefficiencies in each process.
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