Question

A slice of pizza has 500 kcal
. If we could burn the pizza and use all the heat to warm a 50- L
container of cold water, what would be the approximate increase in the temperature of the water? (Note: A liter of cold water weighs about 1 kg
.)

50 ∘C

100 ∘C

5 ∘C

10 ∘C

Answer 1

To determine the increase in temperature of the water when burning a slice of pizza that contains 500 kcal, we can use the specific heat formula:

\[
Q = mc\Delta T
\]

where:
- \( Q \) is the heat energy (in calories, here \( Q = 500 \) kcal),
- \( m \) is the mass of water (in kg, here \( m = 50 \) kg),
- \( c \) is the specific heat capacity of water (approximately \( 1 \) kcal/kg·°C),
- \( \Delta T \) is the change in temperature (in °C).

First, we convert the calories to kcal for consistency:

\[
Q = 500 \text{ kcal} = 500 \times 1000 \text{ cal} = 500,000 \text{ cal}
\]

Now, substituting the values into the equation:

\[
500,000 \text{ cal} = 50 \text{ kg} \times 1 \text{ kcal/kg·°C} \times \Delta T
\]

Since \( 1 \text{ kcal} = 1000 \text{ cal} \), we have:

\[
500,000 \text{ cal} = 50 \text{ kg} \times 1000 \text{ cal/kg·°C} \times \Delta T
\]

Now simplifying:

\[
500,000 = 50,000 \Delta T
\]

Now, solving for \( \Delta T \):

\[
\Delta T = \frac{500,000}{50,000} = 10 \text{ °C}
\]

Thus, the approximate increase in the temperature of the water is \( \Delta T = 10 \text{ °C} \).

Therefore, the answer is **10 °C**.