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
\[ \bbox[#FFCCCC,10px]
{E=50\;\text{g}}
\]