Energy basic question 1
Posted: Wed Nov 06, 2024 9:57 am
An aluminum kettle weighs 1.05 kg.
(a) What is the heat capacity of the kettle?
(b) How much heat is required to increase the temperature of this kettle from 23.0 °C to 99.0 °C?
(c) How much heat is required to heat this kettle from 23.0 °C to 99.0 °C if it contains 1.25 L of water
(density of 0.997 g/mL and a specific heat of 4.184 J/g °C)?
Answer:
Let's break down each part of the problem:
### (a) Heat Capacity of the Kettle
The heat capacity (\(C\)) of an object is given by:
\[ C = m \cdot c \]
where:
- \(m\) is the mass of the object
- \(c\) is the specific heat capacity of the material
For aluminum:
- Mass (\(m\)) = 1.05 kg
- Specific heat capacity (\(c\)) = 0.897 J/g°C = 897 J/kg°C⁸
So, the heat capacity of the kettle is:
\[ C = 1.05 \, \text{kg} \times 897 \, \text{J/kg°C} = 941.85 \, \text{J/°C} \]
### (b) Heat Required to Increase Temperature of the Kettle
The heat (\(Q\)) required to change the temperature is given by:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \(\Delta T\) is the change in temperature
For the kettle:
- Initial temperature = 23.0 °C
- Final temperature = 99.0 °C
- \(\Delta T = 99.0 \, \text{°C} - 23.0 \, \text{°C} = 76.0 \, \text{°C}\)
So, the heat required is:
\[ Q = 1.05 \, \text{kg} \times 897 \, \text{J/kg°C} \times 76.0 \, \text{°C} = 71,583.6 \, \text{J} \]
### (c) Heat Required to Heat the Kettle and Water
First, calculate the heat required to heat the water.
For water:
- Volume = 1.25 L = 1250 mL
- Density = 0.997 g/mL
- Mass (\(m\)) = 1250 mL \times 0.997 g/mL = 1246.25 g = 1.24625 kg
- Specific heat capacity (\(c\)) = 4.184 J/g°C = 4184 J/kg°C³
So, the heat required to heat the water is:
\[ Q_{\text{water}} = 1.24625 \, \text{kg} \times 4184 \, \text{J/kg°C} \times 76.0 \, \text{°C} = 396,318.72 \, \text{J} \]
Now, add the heat required for the kettle and the water:
\[ Q_{\text{total}} = Q_{\text{kettle}} + Q_{\text{water}} = 71,583.6 \, \text{J} + 396,318.72 \, \text{J} = 467,902.32 \, \text{J} \]
### Summary
- (a) Heat capacity of the kettle: \(941.85 \, \text{J/°C}\)
- (b) Heat required to increase the temperature of the kettle: \(71,583.6 \, \text{J}\)
- (c) Heat required to heat the kettle and water: \(467,902.32 \, \text{J}\)
.
(a) What is the heat capacity of the kettle?
(b) How much heat is required to increase the temperature of this kettle from 23.0 °C to 99.0 °C?
(c) How much heat is required to heat this kettle from 23.0 °C to 99.0 °C if it contains 1.25 L of water
(density of 0.997 g/mL and a specific heat of 4.184 J/g °C)?
Answer:
Let's break down each part of the problem:
### (a) Heat Capacity of the Kettle
The heat capacity (\(C\)) of an object is given by:
\[ C = m \cdot c \]
where:
- \(m\) is the mass of the object
- \(c\) is the specific heat capacity of the material
For aluminum:
- Mass (\(m\)) = 1.05 kg
- Specific heat capacity (\(c\)) = 0.897 J/g°C = 897 J/kg°C⁸
So, the heat capacity of the kettle is:
\[ C = 1.05 \, \text{kg} \times 897 \, \text{J/kg°C} = 941.85 \, \text{J/°C} \]
### (b) Heat Required to Increase Temperature of the Kettle
The heat (\(Q\)) required to change the temperature is given by:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \(\Delta T\) is the change in temperature
For the kettle:
- Initial temperature = 23.0 °C
- Final temperature = 99.0 °C
- \(\Delta T = 99.0 \, \text{°C} - 23.0 \, \text{°C} = 76.0 \, \text{°C}\)
So, the heat required is:
\[ Q = 1.05 \, \text{kg} \times 897 \, \text{J/kg°C} \times 76.0 \, \text{°C} = 71,583.6 \, \text{J} \]
### (c) Heat Required to Heat the Kettle and Water
First, calculate the heat required to heat the water.
For water:
- Volume = 1.25 L = 1250 mL
- Density = 0.997 g/mL
- Mass (\(m\)) = 1250 mL \times 0.997 g/mL = 1246.25 g = 1.24625 kg
- Specific heat capacity (\(c\)) = 4.184 J/g°C = 4184 J/kg°C³
So, the heat required to heat the water is:
\[ Q_{\text{water}} = 1.24625 \, \text{kg} \times 4184 \, \text{J/kg°C} \times 76.0 \, \text{°C} = 396,318.72 \, \text{J} \]
Now, add the heat required for the kettle and the water:
\[ Q_{\text{total}} = Q_{\text{kettle}} + Q_{\text{water}} = 71,583.6 \, \text{J} + 396,318.72 \, \text{J} = 467,902.32 \, \text{J} \]
### Summary
- (a) Heat capacity of the kettle: \(941.85 \, \text{J/°C}\)
- (b) Heat required to increase the temperature of the kettle: \(71,583.6 \, \text{J}\)
- (c) Heat required to heat the kettle and water: \(467,902.32 \, \text{J}\)
.