   Chapter 12, Problem 74AP

Chapter
Section
Textbook Problem

Hydrothermal vents deep on the ocean floor spout water at temperatures as high as 570°C. This temperature is below the boiling point of water because of the immense pressure at that depth. Because the surrounding ocean temperature is at 4.0°C, an organism could use the temperature gradient as a source of energy. (a) Assuming the specific heat of water under these conditions is 1.0 cal/g · °C, how much energy is released when 1.0 L of water is cooled from 570°C to 4.0°C? (b) What is the maximum usable energy an organism can extract from this energy source? (Assume the organism has some internal type of heat engine acting between the two temperature extremes.) (c) Water from these vents contains hydrogen sulfide (H2S) at a concentration of 0.90 mmole/L. Oxidation of 1.0 mole of H2S produces 310 kJ of energy. How much energy is available through H2S oxidation of 1.0 L of water?

(a)

To determine
The energy released from water when it cools.

Explanation

Given Info:

High temperature of the water is 570°C .

Low temperature of the water is 4.0°C .

Specific heat of the water is 1.0cal/g°C .

The amount of water cooled is 1.0L

Formula to calculate the energy released from water is,

Qh=mc(ΔT) (I)

• ΔT is the change in temperature
• m is the mass of the water
• c is the specific heat

Since, m=ρV

Equation (I) gives,

Qh=(ρV)c(ΔT)

• ρ is the density of water
• V is the volume of water

Substitute 566°C for ΔT , 1.0cal/g°C for c, 1g/cm3 for ρ and 1

(b)

To determine
The maximum usable energy an organism can extract from this energy source.

(c)

To determine
The energy available through H2S oxidation of 1.0L of water.

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