How many kilojoules of heat are absorbed when 1.00L of water is heated from 18 degrees C to 85 degrees C?
Can you please explain how to do this problem? I do not understand what to do, or understand the formula.
q = mass x specific heat x delta T.
q = the heat absorbed.
mass = mass of the water. That is listed in the formula. 1.0 L H2O has a mass of 1000 g.
specific heat. Do you know the specific heat of water. It goes here.
delta T is the difference in temperature, in this case it is 85-18.
230328
To calculate the amount of heat absorbed when heating water, you need to use the formula:
q = m * C * ΔT
where:
q is the amount of heat (in joules),
m is the mass of the substance (in grams),
C is the specific heat capacity of the substance (in J/g·°C), and
ΔT is the change in temperature (in °C).
To solve this problem, follow these steps:
Step 1: Convert the volume of water (1.00L) to mass (m) in grams. To do this, you can use the density of water, which is approximately 1 g/mL:
m = density * volume
m = 1 g/mL * 1.00 L
m = 1000 g
So, the mass of the water is 1000 grams.
Step 2: Determine the specific heat capacity (C) of water. The specific heat capacity of water is given as C = 4.184 J/g·°C.
Step 3: Calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 85°C - 18°C
ΔT = 67°C
Step 4: Now, plug the values into the formula to calculate the amount of heat (q) absorbed:
q = m * C * ΔT
q = 1000 g * 4.184 J/g·°C * 67°C
Calculating this equation will give you the answer in joules. However, the question asks for the answer in kilojoules, so you need to convert the answer to kilojoules by dividing by 1000:
q = (1000 g * 4.184 J/g·°C * 67°C) / 1000
q ≈ 281,372 J / 1000
q ≈ 281.37 kJ
Therefore, when 1.00L of water is heated from 18°C to 85°C, approximately 281.37 kilojoules of heat are absorbed.