To calculate the amount of heat that must be added to warm 32.5 grams of brick from 22.0°C to 40.4°C, we will use the concept of specific heat capacity and the formula:
[tex]\[ Q = mc\Delta T \][/tex]
where:
- \( Q \) is the amount of heat added (in calories),
- \( m \) is the mass of the substance (in grams),
- \( c \) is the specific heat capacity of the substance (in \(\frac{cal}{g \cdot °C}\)),
- \( \Delta T \) is the change in temperature (in °C).
Given:
- Mass \( m = 32.5 \text{ grams} \),
- Initial temperature \( T_i = 22.0 \text{°C} \),
- Final temperature \( T_f = 40.4 \text{°C} \),
- Specific heat of brick \( c = 0.20 \frac{ cal }{ g \cdot °C} \).
First, we calculate the change in temperature (\( \Delta T \)):
[tex]\[ \Delta T = T_f - T_i \][/tex]
Substituting the given values:
[tex]\[ \Delta T = 40.4 \text{°C} - 22.0 \text{°C} \][/tex]
[tex]\[ \Delta T = 18.4 \text{°C} \][/tex]
Next, we apply the formula \( Q = mc\Delta T \) to find the amount of heat added:
[tex]\[ Q = 32.5 \text{ g} \times 0.20 \frac{ cal }{ g \cdot °C} \times 18.4 \text{°C} \][/tex]
Now, we multiply these values together:
[tex]\[ Q = 32.5 \times 0.20 \times 18.4 \][/tex]
[tex]\[ Q = 119.6 \text{ calories} \][/tex]
Therefore, the amount of heat that must be added to warm 32.5 grams of brick from 22.0°C to 40.4°C is 119.6 calories.
heat added: [tex]\( \boxed{119.6} \text{ calories} \)[/tex]
