how many joules of heat are needed to heat 900.0g water from 18.0 C to 98.0 C?
Well, if I were a clown performing a heat-themed comedy routine, I'd say you're looking to give 900.0 grams of water a little warm makeover, from a chilly 18.0°C to a toasty 98.0°C. To pull off this magical transformation, we need to calculate the amount of heat required.
Now, the specific heat capacity of water is about 4.184 joules/gram°C. So, to calculate the heat, we use the formula:
Q = m * c * ΔT
where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Plugging in the numbers:
Q = 900.0 g * 4.184 J/g°C * (98.0°C - 18.0°C)
Calculating that, we get:
Q ≈ 296,496 joules
So, approximately 296,496 joules of heat are required to give your water that warm and cozy feeling from 18.0°C to 98.0°C.
To calculate the amount of heat needed to heat water, we can use the formula:
Q = m * C * ΔT
where:
Q is the heat (in joules),
m is the mass of the water (in grams),
C is the specific heat capacity of water (4.18 J/g°C), and
ΔT is the change in temperature (in °C).
Given:
m = 900.0 g
ΔT = (98.0°C - 18.0°C) = 80.0°C
Plugging in these values into the formula, we get:
Q = 900.0 g * 4.18 J/g°C * 80.0°C
Simplifying the equation, we have:
Q = 301,680 J
Therefore, the amount of heat needed to heat 900.0 g of water from 18.0°C to 98.0°C is 301,680 joules.
To calculate the amount of heat energy needed to heat a substance, you can use the equation:
Q = mcΔT
Where:
- Q is the amount of heat energy (in joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in joules per gram-degree Celsius)
- ΔT is the change in temperature (in degrees Celsius)
For water, the specific heat capacity (c) is approximately 4.184 J/g°C.
Now let's plug in the values:
m = 900.0 g (given)
c = 4.184 J/g°C (specific heat capacity of water)
ΔT = (final temperature - initial temperature) = (98.0°C - 18.0°C) = 80.0°C
Substituting these values into the equation:
Q = (900.0 g) × (4.184 J/g°C) × (80.0°C)
Calculating this:
Q = 301,968 joules
Therefore, approximately 301,968 joules of heat are needed to heat 900.0g of water from 18.0°C to 98.0°C.