how much heat is needed to convert 30 g of water at 10 C to 110 C
q1 = heat needed to raise T of water from 10 to 100.
q1 = mass x specific heat liquid water x (Tfinal-Tintial)
q2 = heat needed to vaporize the liquid water at 100 C to steam at 100 C.
q2 = mass water x heat vaporization.
q3 = heat needed to raise T of steam at 100 C to steam at 110 C.
q3 = mass steam x specific heat steam x (Tfinal-Tinitial)
Total Q = q1 + q2 + q3.
To calculate the amount of heat required to convert a given mass of a substance from one temperature to another, you need to know the specific heat capacity of the substance.
The specific heat capacity (also known as specific heat) is a property that represents the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin).
For water, the specific heat capacity is approximately 4.184 J/g·°C.
To find the amount of heat required to raise the temperature of 30 g of water from 10 °C to 110 °C, you can use the formula:
Q = mcΔT
where:
Q is the heat energy (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity (in J/g·°C), and
ΔT is the change in temperature (in °C).
Substituting the given values into the formula:
Q = (30 g) × (4.184 J/g·°C) × (110 °C - 10 °C)
Q = (30 g) × (4.184 J/g·°C) × (100 °C)
Q = 12552 J
Therefore, approximately 12,552 joules of heat energy is needed to convert 30 g of water at 10 °C to 110 °C.