Question

2.71 grams of a fuel was burned releasing 4.10 kJ of energy into some amount of water. The water changed from 22.2°C to 53.1°C. How much water was heated?

Answer 1

In this case, we have Q = 4.10 kJ, m is unknown, c is the specific heat capacity of water (4.184 J/g °C), ΔT = 53.1 °C - 22.2 °C = 30.9 °C. Approximately 97.82 grams of water was heated.

Step-by-step explanation:

To find the amount of water heated, we can use the formula:

Q = mcΔT

Where Q is the heat energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

In this case, we have Q = 4.10 kJ, m is unknown, c is the specific heat capacity of water (4.184 J/g °C), ΔT = 53.1 °C - 22.2 °C = 30.9 °C.

Rearranging the formula to solve for m:
m = Q / (c ΔT)

Substituting the given values:

m = 4.10 kJ / (4.184 J/g °C * 30.9 °C)

m = 97.82 g

Therefore, approximately 97.82 grams of water was heated.