Exactly 11.5 mL of water at 34.0 °C is added to a hot iron skillet.

Question

All of the water is converted into steam at 100.0°C. The mass of the pan is 1.10 kg and the molar heat capacity of iron is 25.19 J/(mol·°C).

What is the temperature change of the skillet?

Answer 1

First convert 11.5 mL of water to a mass: 1 mL = 1 cm³, and density of water is 1 g/cm³, so m = 11.5 g = 0.0115 kg. To heat this much water to the boiling point, Q = m c ΔT = (0.0115 kg) (4186 J/kg °C) (100 - 34) = 3,177 J. Then to convert the hot water to steam, we use the latent heat of vaporization Q = m L_v = (0.0115 kg) (2.26x10⁶ J/kg) = 25,990 J. The total heat to boil 34 °C water, then, is 29,167 J.

The 1.1 kg skillet loses this much heat in the process. Notice that the heat capacity given for iron uses moles, so n = (1,100 g) / (55.85 g/mol) = 19.7 moles. Q = -29,167 = (19.7 mol) (25.19 J/mol °C) ΔT, ΔT = -58.8 °C.