Question

If we are trying to heat 90 g of water to a boil starting at 22 degrees C, how much energy in joules will we need? The 'c' value for water is 4.18 J/g°C.

A.) 8000 J
B.) 1932 J
C.) 433 J
D.) 8276 J

Answer 1

The correct amount of energy required to heat 90 g of water from 22 degrees C to boiling exceeds the highest option given, which is 8276 J, indicating a possible misprint in the answer choices.

Step-by-step explanation:

Calculating the Energy Required to Heat Water

To determine the amount of energy needed to heat 90 g of water from 22 degrees C to boiling, we can use the equation for heat energy involving specific heat capacity:

Q = mcΔT, where:

• 
  
• The given specific heat capacity of water is 4.18 J/g°C. To bring water to a boil, we assume we need to reach 100 degrees C. This means ΔT is 100°C - 22°C = 78°C.

Therefore, the heat energy required (Q) is calculated as follows:

Q = (90 g) × (4.18 J/g°C) × (78°C)

Q = 29358.6 J

Thus, Q is significantly higher than the highest option given (D) 8276 J, indicating a possible misprint in the provided options. The correct amount of energy in joules required to heat 90 g of water from 22 degrees C to a boil exceeds any of the provided choices.