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