Posted by **Anonymous** on Thursday, November 27, 2008 at 8:22pm.

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.

- physics -
**DrBob222**, Thursday, November 27, 2008 at 9:15pm
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.

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