PLEASE HELP! What amount of water could be heated from 20 degrees C to 50 degrees C by 714 kJ of energy?
q = mass water x specific heat water x delta T.
714,000 = mass water x specific heat water x 30. Substitute specific heat water and solve for mass water.
To determine the amount of water that could be heated by 714 kJ of energy, we need to use the following formula:
q = m * c * ΔT
Where:
q: heat energy transferred (in joules)
m: mass of the substance (in grams)
c: specific heat capacity of the substance (in J/g·°C)
ΔT: change in temperature (in °C)
In this case, we know the heat energy (q) is 714 kJ, the initial temperature (20°C), and the final temperature (50°C). However, we need to find the mass of the water (m) in order to solve for it.
To find the mass of water, we can use the specific heat capacity of water, which is approximately 4.184 J/g·°C.
Let's plug in the values into the equation and solve for the mass (m):
714,000 J = m * 4.184 J/g·°C * (50°C - 20°C)
First, let's calculate the temperature difference (ΔT):
ΔT = 50°C - 20°C = 30°C
Next, rearrange the formula to solve for the mass (m):
m = q / (c * ΔT)
m = 714,000 J / (4.184 J/g·°C * 30°C)
m = 714,000 J / 125.52 J/g
m ≈ 5689.19 g
Therefore, the mass of water that could be heated from 20 °C to 50 °C by 714 kJ of energy is approximately 5689.19 grams.