If the temperature of 150.0g of water increased from 23.3C to 44.8C, how much heat was absorbed
In units of joules. And in units of calories?
q = mass H2O x specific heat H2O x delta T.
Use 4.184 J/g for specific heat H2O for q in joules.
Use 1 cal/g for specific heat H2O for q in calories.
OR a quicker way is to calculate calories first and multiply that answer by 4.184 to convert to J.
To find out how much heat was absorbed by the water, we can use the formula:
Q = m * c * ∆T
where:
- Q is the heat absorbed or released (in joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in J/g°C)
- ∆T is the change in temperature (in °C)
In this case, the mass of water (m) is given as 150.0 grams, the specific heat capacity of water (c) is 4.18 J/g°C, and the change in temperature (∆T) is (44.8°C - 23.3°C) = 21.5°C.
Now, let's plug these values into the equation:
Q = 150.0 g * 4.18 J/g°C * 21.5°C
Calculating this equation gives us:
Q = 135,315 J
Therefore, 150.0 grams of water absorbed 135,315 joules of heat when its temperature increased from 23.3°C to 44.8°C.