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?
5E5cal=50kg*80cal/g*deltaT
deltaTemp=5E4/(5E4*80) deg C
check those numbers. 50 liters is a lot of water.
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?
To find the approximate increase in the temperature of the water when the pizza is burned, we need to use the concept of specific heat. Specific heat is the amount of heat energy required to raise the temperature of a given mass of a substance by a certain amount.
In this case, we need to know the specific heat of water, which is approximately 4.18 Joules per gram per degree Celsius (J/g°C). We also need to convert the kilocalories (kcal) of energy released by the pizza into joules, as specific heat is usually expressed in joules.
To convert kcal to joules, we need to multiply the kcal value by 4,184 Joules per kilocalorie. So, for a 500 kcal slice of pizza, the energy released would be 500 kcal * 4,184 J/kcal = 2,092,000 joules.
Now, we need to calculate the mass of the water in the 50-L container. The density of water is approximately 1 g/mL, so the mass of 50 L (50,000 mL) of water is 50,000 g.
Finally, we can use the equation:
ΔQ = m * c * ΔT
where ΔQ is the heat energy transferred, m is the mass of the substance (water in this case), c is the specific heat of the substance, and ΔT is the change in temperature.
Rearranging the equation to solve for ΔT, we have:
ΔT = ΔQ / (m * c)
Plugging in the values, we get:
ΔT = 2,092,000 J / (50,000 g * 4.18 J/g°C)
ΔT = 9.98°C
Therefore, the approximate increase in the temperature of the water would be approximately 9.98 degrees Celsius.