Solved Problem on Heat
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The block in the figure is made up of equal masses of two substances, A and B, of specific heats cA = 0.20 cal/g°C and cB = 0.30 cal/g°C and the mass of the block is equal to 200 g. Determine:
a) The Heat capacity of the block;
b) The amount of heat that must be supplied to the block so that its temperature rises by 20 ºC;
c) What is the water equivalent of the block?


Problem data:
  • Mass of block:    m = 200 g;
  • Specific heat of substance A:    cA = 0.20 cal/gºC;
  • Specific heat of substance B:    cB = 0.30 cal/gºC.
Solution

a) The heat capacity is given by
\[ \bbox[#99CCFF,10px] {C=m c} \]
the specific heats of each part are known, the mass m of the block is known, and each part has the same mass
\[ m_{A}=m_{B}=100\;\text{g} \]
the heat capacity of each part will then be given by
\[ C_{A}=m_{A}c_{A} \]
\[ C_{B}=m_{B}c_{B} \]
The total heat capacity will be the sum of the heat capacity of each part
\[ \begin{gather} C_{T}=C_{A}+C_{B}\\ C_{T}=m_{A}c_{A}+m_{B}c_{B} \end{gather} \]
substituting the data
\[ \begin{gather} C_{T}=100\times 0.20+100\times 0.30\\ C_{T}=20+30 \end{gather} \]
\[ \bbox[#FFCCCC,10px] {C_{T}=50\;\text{cal/°C}} \]

b) The heat capacity given as a function of the amount of heat and temperature is given by
\[ \bbox[#99CCFF,10px] {C=\frac{Q}{\Delta t}} \]
then the amount of heat can be calculated as
\[ Q=C\Delta t \]
as we want an increase of 20 °C this will be the value for Δt, using the value of the total heat capacity of the block calculated in the previous item
\[ Q=50\times 20 \]
\[ \bbox[#FFCCCC,10px] {Q=1000\;\text{cal}} \]

c) The water equivalent (given in grams) is numerically equal to the heat capacity (given in calories per degree Celsius)
\[ E(\text{g})\overset{\text{N}}{=}C(\text{cal/°C}) \]
The water equivalent will be
\[ \bbox[#FFCCCC,10px] {E=50\;\text{g}} \]
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