Exactly 19.5 mL of water at 27.0 °C is added to a hot
iron skillet. All of the water is converted to steam at 100.0°C.
The mass of the skillet is 1.45 kg. What is the change
in temperature of the skillet?
Answer should be in degrees Celsius
Q = m Cp dT
Q = 19.5 x 4.184 x (100-27)
Q1 = 5956 J
Q2 = heat of vaporization x mass
= 2260 x 19.5
= 44070 J
total heat = Q = Q1 + Q2
Q = 50026 J
heat lost by water = heat gain by skillet
50026 = 1450 x 0.444 x dT
dT = 77.7 oC
change in temperature of the skillet = 77.7 oC
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