Solved Problem on Heat
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From what height should fall a mass of water so that its final energy, converted into heat, increase the temperature of this mass of 1 °C? Assume that there is no energy loss in the system.
Data: 1 cal = 4.18 J, g = 9.8 m/s2, c = 1 cal/g°C.


Problem data:
  • Chabge in water temperature:    Δ t = 1 °C;
  • Specific heat of water:    c = 1 cal/g°C;
  • Mechanical equivalent of heat:    1 cal = 4.18 J;
  • Acceleration due to gravity:    g = 9.8 m/s2.
Problem diagram:

In the initial position, at a height of h, all the mechanical energy of the water mass is in the form of potential energy EP. When the mass reaches the ground, assuming that there is no energy loss, all potential energy becomes heat Q (Figure 1) that heats the water.
Figure 1
Solution

First, we will convert the specific heat from the water given in calories per gram-degree Celsius (cal/g °C) to joules per kilogram-degree Celsius (J/kg °C) used in the International System of Units (S.I.)
\[ c=1\;\frac{\cancel{\text{cal}}}{\cancel{\text{g}}\text{°C}}\times\frac{4.18\;\text{J}}{1\;\cancel{\text{cal}}}\times\frac{1000\;\cancel{\text{g}}}{1\;\text{kg}}=4180\;\frac{\text{J}}{\text{kg°C}} \]
The Potential Energy is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {U=mgh} \tag{I} \end{gather} \]
The equation of heat received by the body is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {Q=mc\Delta t} \tag{II} \end{gather} \]
The initial potential energy is converted into heat at the end, equating expressions (I) and (II)
\[ \begin{gathered} U=Q\\ \cancel{m}gh=\cancel{m}c\Delta t\\ h=\frac{c \Delta t}{g} \end{gathered} \]
substituting the problem data
\[ h=\frac{4180\times 1}{9.8} \]
\[ \bbox[#FFCCCC,10px] {h=426.5\;\text{m}} \]
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