Answer :
To determine the amount of heat that must be added to raise the temperature of a cup of coffee, we need to use the formula for heat transfer in calories:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat energy added (in calories),
- [tex]\( m \)[/tex] is the mass of the coffee (in grams),
- [tex]\( c \)[/tex] is the specific heat capacity (in cal/g.°C),
- [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).
Given the data:
- Initial temperature ([tex]\( T_{\text{initial}} \)[/tex]) = 20.5°C
- Final temperature ([tex]\( T_{\text{final}} \)[/tex]) = 95.6°C
- Volume of coffee = 50.0 mL
- Density of coffee = 1.00 g/mL
- Specific heat capacity ([tex]\( c \)[/tex]) = 1 cal/g.°C
First, we determine the mass of the coffee, which is equal to its volume times its density:
[tex]\[ m = 50.0 \, \text{mL} \times 1.00 \, \text{g/mL} = 50.0 \, \text{g} \][/tex]
Next, we calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \][/tex]
[tex]\[ \Delta T = 95.6°C - 20.5°C \][/tex]
[tex]\[ \Delta T = 75.1°C \][/tex]
Now, we can calculate the heat energy added ([tex]\( Q \)[/tex]) using the formula:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
[tex]\[ Q = 50.0 \, \text{g} \times 1 \, \text{cal/g.°C} \times 75.1°C \][/tex]
[tex]\[ Q = 3754.9999999999995 \, \text{cal} \][/tex]
Finally, we convert the heat energy from calories to kilocalories (since 1 kcal = 1000 cal):
[tex]\[ Q_{\text{kcal}} = \frac{Q}{1000} \][/tex]
[tex]\[ Q_{\text{kcal}} = \frac{3754.9999999999995 \, \text{cal}}{1000} \][/tex]
[tex]\[ Q_{\text{kcal}} = 3.7549999999999994 \, \text{kcal} \][/tex]
Therefore, the amount of heat that must be added to raise the temperature of the cup of coffee from 20.5°C to 95.6°C is approximately 3.755 kcal.
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat energy added (in calories),
- [tex]\( m \)[/tex] is the mass of the coffee (in grams),
- [tex]\( c \)[/tex] is the specific heat capacity (in cal/g.°C),
- [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).
Given the data:
- Initial temperature ([tex]\( T_{\text{initial}} \)[/tex]) = 20.5°C
- Final temperature ([tex]\( T_{\text{final}} \)[/tex]) = 95.6°C
- Volume of coffee = 50.0 mL
- Density of coffee = 1.00 g/mL
- Specific heat capacity ([tex]\( c \)[/tex]) = 1 cal/g.°C
First, we determine the mass of the coffee, which is equal to its volume times its density:
[tex]\[ m = 50.0 \, \text{mL} \times 1.00 \, \text{g/mL} = 50.0 \, \text{g} \][/tex]
Next, we calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \][/tex]
[tex]\[ \Delta T = 95.6°C - 20.5°C \][/tex]
[tex]\[ \Delta T = 75.1°C \][/tex]
Now, we can calculate the heat energy added ([tex]\( Q \)[/tex]) using the formula:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
[tex]\[ Q = 50.0 \, \text{g} \times 1 \, \text{cal/g.°C} \times 75.1°C \][/tex]
[tex]\[ Q = 3754.9999999999995 \, \text{cal} \][/tex]
Finally, we convert the heat energy from calories to kilocalories (since 1 kcal = 1000 cal):
[tex]\[ Q_{\text{kcal}} = \frac{Q}{1000} \][/tex]
[tex]\[ Q_{\text{kcal}} = \frac{3754.9999999999995 \, \text{cal}}{1000} \][/tex]
[tex]\[ Q_{\text{kcal}} = 3.7549999999999994 \, \text{kcal} \][/tex]
Therefore, the amount of heat that must be added to raise the temperature of the cup of coffee from 20.5°C to 95.6°C is approximately 3.755 kcal.
