how many kilojoules of heat are absorbed when 1.00L of water is heated from 18 degrees celsius to 85 degrees celsius?

Well, let's do some calculations! But first, let me ask you, did you know that water has a fantastic sense of humor? It always has the perfect response, no matter the situation! So, let's see how many kilojoules of heat are absorbed when your hilarious water decides to heat up!

To solve this problem, we need to use the equation:

Q = mcΔT

Where Q is the heat energy (in joules), m is the mass (in grams), c is the specific heat capacity of water (which is 4.18 J/g°C), and ΔT is the change in temperature (in degrees Celsius).

Now, since you're giving me the volume of water instead of mass, no worries, I'll convert the volume to mass using the density of water, which is approximately 1 gram/mL.

So, 1.00L of water is equivalent to 1000 grams. And the change in temperature is:

ΔT = (85°C - 18°C) = 67°C

Now that we have all the values, let's plug them into the equation! Drumroll please...

Q = (1000g) * (4.18 J/g°C) * (67°C)

Calculating...

Q = 281,060 joules

But you asked for kilojoules, so let's convert it! Ta-da!

Q = 281.06 kilojoules

So, when your water is feeling particularly heated, it absorbs approximately 281.06 kilojoules of heat! Isn't water just a cool comedian?

To calculate the amount of heat absorbed, you can use the formula:

Q = m * c * ΔT

Where:
Q is the amount of heat absorbed (in joules)
m is the mass of the substance (in kilograms)
c is the specific heat capacity of the substance (in J/g·°C or J/kg·°C)
ΔT is the change in temperature (in °C)

First, let's calculate the mass of water in the given volume of 1.00L. The density of water is approximately 1g/mL, so the mass of 1.00L of water is:

Mass = Volume * Density = 1.00L * 1000g/L = 1000g

Next, we need to convert the mass to kilograms:

Mass = 1000g * (1kg/1000g) = 1kg

The specific heat capacity of water is approximately 4.18 J/g·°C (or 4186 J/kg·°C).

Now we can plug the values into the formula:

Q = m * c * ΔT
Q = 1kg * 4.18 J/g·°C * (85°C - 18°C)

ΔT = 85°C - 18°C = 67°C

Q = 1kg * 4.18 J/g·°C * 67°C
Q = 282.2 J/g * 67°C
Q = 18,911.4 J

To convert joules to kilojoules, divide the answer by 1000:

Q = 18,911.4 J / 1000 = 18.91 kJ

Therefore, approximately 18.91 kilojoules of heat are absorbed when 1.00L of water is heated from 18°C to 85°C.

To calculate the amount of heat absorbed by the water, we can use the equation:

q = mcΔT,

where q represents the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

First, let's calculate the mass of the water in grams. The density of water is approximately 1 g/mL or 1 g/cm³. Since we have 1.00 L of water, which is equivalent to 1000 mL or 1000 cm³:

mass = volume × density = 1000 cm³ × 1 g/cm³ = 1000 g.

Next, let's calculate the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/(g·°C) or 4.18 J/(g·K). Since we are working with kilojoules, we will convert this value to kilojoules:

specific heat capacity = 4.18 J/(g·°C) × 0.001 kJ/J = 0.00418 kJ/(g·°C).

Now, we can calculate the change in temperature:

ΔT = final temperature - initial temperature = 85°C - 18°C = 67°C.

Finally, we can substitute these values in the equation to find the amount of heat absorbed:

q = (mass) × (specific heat capacity) × (ΔT)
= (1000 g) × (0.00418 kJ/(g·°C)) × (67°C)
= 279.86 kJ.

Therefore, approximately 279.86 kilojoules of heat are absorbed when 1.00 L of water is heated from 18 degrees Celsius to 85 degrees Celsius.

The mass of 1 liter of water is very close to 1 kg

q = (specific heat)*(mass)*deltaT
specific heat of water = 4.18kJ/kg.C
q = (4.18kJ/kg.C)(1.00kg)(85ºC-18ºC)