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Chapter 14: Problem 77

If \(4.0 \mathrm{g}\) of steam at \(100.0^{\circ} \mathrm{C}\) condenses to water on a burn victim's skin and cools to \(45.0^{\circ} \mathrm{C},\) (a) how much heat is given up by the steam? (b) If the skin was originally at $37.0^{\circ} \mathrm{C},$ how much tissue mass was involved in cooling the steam to water? See Table 14.1 for the specific heat of human tissue.

Short Answer

Expert verified
Answer: The steam gives up 9960.74 J of heat during condensation and cooling to 45.0°C. The mass of the tissue involved in cooling the steam to water is 0.358 kg.

Step by step solution

01

Calculate the heat released during condensation

To calculate the heat released during condensation, we will use the Q = mL formula, where Q is the heat released, m is the mass of the steam, and L is the latent heat of vaporization. The latent heat of vaporization for water is 2.257 x 10^6 J/kg. Q_condensation = m * L Q_condensation = 4.0 * 10^(-3) kg * 2.257 x 10^(6) J/kg = 9040 J
02

Calculate the heat released as the water cools to 45.0°C

We will now calculate the heat released as the water cools down to 45.0°C. For this, we will use the Q = mcΔT formula, where Q is the heat released, m is the mass of the water, c is the specific heat capacity of water (4.186 x 10^3 J/kg·°C), and ΔT is the change in temperature. ΔT = 100.0 - 45.0 = 55.0°C Q_cooling = m * c * ΔT Q_cooling = 4.0 * 10^(-3) kg * 4.186 x 10^3 J/kg·°C * 55.0°C = 920.74 J
03

Calculate the total heat released by the steam

Now, we will add the heat released during condensation and the heat released during cooling to find the total heat given up by the steam. Q_total = Q_condensation + Q_cooling Q_total = 9040 J + 920.74 J = 9960.74 J The steam gives up 9960.74 J of heat. #b) How much tissue mass was involved in cooling the steam to water?#
04

Calculate the heat absorbed by the tissue

Since the heat given up by the steam should equal the heat absorbed by the human tissue, we already know the heat absorbed by the tissue is 9960.74 J from part (a).
05

Calculate the mass of the tissue

Now, we will use the Q = mcΔT equation, along with the specific heat of human tissue to find the mass of the tissue. The specific heat of human tissue is 3.47 x 10^3 J/kg·°C, as provided in Table 14.1. ΔT = 45.0 - 37.0 = 8.0°C Using the equation Q = mcΔT, we can solve for the mass (m): m = Q / (c * ΔT) m = 9960.74 J / (3.47 x 10^3 J/kg·°C * 8.0°C) = 0.358 kg The mass of the tissue involved in cooling the steam to water is 0.358 kg.

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