Final answer:
The heat transferred to the skin by 25.0 grams of steam is calculated using the latent heat of vaporization, resulting in 56,400 Joules (or 56.4 kJ).
Step-by-step explanation:
To calculate the amount of heat (Q2) transferred to the skin by 25.0 g of steam, we use the formula: Q = mLv, where m is the mass of the steam and Lv is the latent heat of vaporization. Given that the heat of vaporization for steam is L=2.256×106 J/kg, we first need to convert the mass of the steam to kilograms by dividing 25.0 g by 1000, which gives us 0.025 kg. Next, we multiply the mass (0.025 kg) by the latent heat of vaporization (2.256×106 J/kg) to find the total heat transferred:
Q = mLv = (0.025 kg)(2.256 × 106 J/kg) = 56,400 J or 56.4 kJ.
Therefore, the heat transferred to the skin by the steam is 56,400 Joules (or 56.4 kJ).