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