posted by Anonymous on .
The amount of heat Q needed to turn a mass m of room temperature T_1 water into steam at 100degrees C T_2 can be found using the specific heat c of water and the heat of vaporization H_v of water at 1 atmosphere of pressure.
Suppose that a commercial espresso machine in a coffee shop turns 1.50 kg of water at 22.0degreesC into steam at 100degreesC. If c = 4187 and H_v=2258 kJ/kg, how much heat Q is absorbed by the water from the heating resistor inside the machine?
Assume that this is a closed and isolated system.
q1 = heat needed to heat 1.5 kg water from room T to 100 C.
q1 = mass x specific heat x (Tfinal-Tinitial).
q1 = 1.5 kg x 4187 J/kg*C x (100-22) = ??
q2 = heat needed to turn 1.5 kg water at 100 C to steam at 100 C = mass x Hvap
q2 = 1.5 kg x 2,258,000 J/kg = ??
Total heat is q1 + q2.
Check my thinking.
yeah this is right, I used this logic to answer meh mastering physics problem.
- 8==> (girthydick)