how much heat must be absorbed by a 500 g pot of water to raise the temperature from 20 degree C to 30 degree C?
Well, I don't mean to brag, but water is pretty cool. Pun intended. Now, let me do some quick calculations for you. The specific heat capacity of water is around 4.18 J/g°C, which means that it takes 4.18 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius. So, for your 500 gram pot of water to go from 20°C to 30°C, you would need to absorb around... (calculating) approximately 20,900 joules of heat. But hey, don't worry, I won't leave you high and dry.
To calculate the amount of heat required to raise the temperature of water, we can use the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q = Amount of heat absorbed (in Joules)
m = Mass of the substance (in grams)
c = Specific heat capacity (in J/g°C)
ΔT = Change in temperature (in °C)
Given:
m = 500 g
ΔT = 30°C - 20°C = 10°C
The specific heat capacity of water is approximately 4.18 J/g°C.
Substituting the given values into the formula:
Q = 500 g * 4.18 J/g°C * 10°C
Q = 20,900 Joules
Therefore, approximately 20,900 Joules of heat must be absorbed by the 500 g pot of water to raise its temperature from 20°C to 30°C.
To calculate the heat absorbed by a substance, we can use the equation:
Q = m * c * ΔT
where:
Q is the heat absorbed or released (in joules),
m is the mass of the substance (in kilograms),
c is the specific heat capacity of the substance (in joules per gram-degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
In this case, you have a 500 g pot of water, so the mass (m) will be 500 g, which we must convert to kilograms before substituting it into the equation.
m = 500 g ÷ 1000 = 0.5 kg
The specific heat capacity (c) of water is approximately 4.18 joules per gram-degree Celsius.
Substituting the given values into the equation, we have:
Q = 0.5 kg * 4.18 J/g°C * (30°C - 20°C)
Simplifying further:
ΔT = 30°C - 20°C = 10°C
Q = 0.5 kg * 4.18 J/g°C * 10°C
Now we can calculate the answer:
Q = 20.9 J/g * 10 g = 209 J (joules)
Therefore, approximately 209 joules of heat must be absorbed by the 500 g pot of water to raise its temperature from 20°C to 30°C.
If you don't include the energy that has to be added to the pot, the amount that has to be added to the water is M*C*(T2 - T1) = 5000 calories
C is the specific heat of water,
1.00 cal/(g degC)
You must have seen this formula before, in your text or class notes.