Approximately how many joules of heat are needed to raise the temperature of a 4.0-gram sample of water to 8.0C?
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16.8
Well, well, well, it seems like we've got ourselves a little thermodynamics puzzle here! To solve this, we'll need the specific heat capacity of water, which is approximately 4.18 joules per gram per degree Celsius.
But before we get into the math, let's set the stage. Picture a 4.0-gram sample of water, preparing for the performance of a lifetime - the temperature rise to 8.0 degrees Celsius!
Alright, let's get calculating! We'll multiply the mass of the water (4.0 grams) by the specific heat capacity of water (4.18 joules per gram per degree Celsius) and the temperature change (8.0 degrees Celsius).
So, the joules of heat needed can be calculated as follows:
4.0 grams x 4.18 joules per gram per degree Celsius x 8.0 degrees Celsius = approximately 133.76 joules of heat
There you have it! It takes approximately 133.76 joules of heat to raise the temperature of our water sample. Bravo, water, bravo! Keep up the fantastic work!
To find the amount of heat needed, we can use the specific heat capacity formula:
Q = m * c * ΔT
where:
Q is the heat needed (in Joules)
m is the mass of the water sample (in grams)
c is the specific heat capacity of water (approximately 4.18 J/g°C)
ΔT is the change in temperature (in °C)
Given:
m = 4.0 grams
ΔT = 8.0°C
c = 4.18 J/g°C
Substituting the values into the formula:
Q = 4.0 g * 4.18 J/g°C * 8.0°C
Calculating:
Q = 134.08 Joules (rounded to two decimal places)
Approximately 134.08 Joules of heat are needed to raise the temperature of the 4.0-gram sample of water to 8.0°C.
To calculate the amount of energy required to raise the temperature of a substance, you can use the formula:
q = m * c * ΔT
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,
- ΔT is the change in temperature in Celsius.
For water, the specific heat capacity is approximately 4.18 J/g°C.
Now, let's substitute the given values into the formula:
m = 4.0 g
c = 4.18 J/g°C
ΔT = 8.0°C
q = (4.0 g) * (4.18 J/g°C) * (8.0°C)
q = 134.08 J
Therefore, approximately 134.08 joules of heat are needed to raise the temperature of the 4.0-gram sample of water to 8.0°C.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial).
4
q = 4.0g x 4.184 J/g*C x (8-Tinitial).
Can't do it without Tinitial.
That 8.0 C may be to raise the T BT 8.0 C. in which case Tfinal-Tinitial = 8.0