Heat and the First Law of Thermodynamics

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SUMMARY

The discussion centers on calculating the increase in water temperature using Joule's apparatus, where two 1.50 kg masses fall through a distance of 3.50 m, affecting 190 g of water. The initial calculation incorrectly equated work done to temperature change, leading to an erroneous result of 103°C. The correct relationship is defined by the equation W = c m ΔT, where W is the work done, c is the specific heat of water, m is the mass of the water, and ΔT is the change in temperature.

PREREQUISITES
  • Understanding of gravitational potential energy (PE = mgh)
  • Familiarity with the concept of work in physics
  • Knowledge of specific heat capacity, particularly for water
  • Basic algebra for manipulating equations
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Consider Joule's apparatus described in Figure 20.1. Each of the two masses is 1.50 kg, and the tank is filled with 190 g of water. What is the increase in the temperature of the water after the masses fall through a distance of 3.50 m?

For this I know that PE=W, and that 2mgh is proportional to the change in temperature, but for some reason I can't get the correct answer

2(1.5 g)(9.8 m/s2)(3.5m)=103 C, but this is not correct
 
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What you calculated is the work done in Joules, not the change in temperature. But the temperature change is proportional to the work done, but proportional does not mean equal. The relationship is as follows:
W = c m \Delta T

Where c is the specific heat of water (look it up) and m is the mass of the water.
 
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