a) Object A
When a certain amount of energy Q is supplied to a sample of substance with mass m, the temperature of the substance increases by [tex]\Delta T[/tex], according to the equation :
[tex]Q=mC_s \Delta T[/tex]
where [tex]C_s[/tex] is the specific heat capacity of the substance .
The equation can be rewritten as
[tex]C_s = \frac{Q}{m \Delta T}[/tex]
In our problem, we have two different objecs A and B (we assume they have same mass and they are at same initial temperature). Both objects are placed in a beaker containing 1000 g of water at 10.0 °C. Each object gives off energy to the water, until the object is in thermal equilibrium (=same temperature) with the water. The amount of energy given off by each object is equal to the amount of energy absorbed by the water, so:
[tex]m_w C_w (T_{eq} - T_{wi})=m_o C_o (T_{oi}-T_{eq})[/tex]
where
mw is the mass of water
Cw is the specifi heat of water
Teq is the temperature at equilibrium
Twi is the initial temperature of water
[tex]m_o[/tex] is the mass of the object
[tex]C_o[/tex] is the specific heat capacity of the object
[tex]T_{oi}[/tex] is the initial temperature of the object
Solving for [tex]C_o[/tex],
[tex]C_o=\frac{m_w C_w (T_{eq} - T_{wi})}{m_o(T_{oi}-T_{eq})}[/tex]
For object A, the increase in temperature of the water is
[tex]T_{eq}-T_{wi}=3.50^{\circ}[/tex]
So the formula becomes
[tex]C_A=\frac{3.5 m_w C_w}{m_o(T_{oi}-T_{eq})}[/tex]
For object B, the increase in temperature of the water is
[tex]T_{eq}-T_{wi}=2.60^{\circ}[/tex]
So the formula becomes
[tex]C_B=\frac{2.6 m_w C_w}{m_o(T_{oi}-T_{eq})}[/tex]
Also, we have to notice that if the two objects start at the same temperature, then the equilibrium temperature in case A) is higher than in case B, therefore the denominator [tex]T_{oi}-T_{eq}[/tex] is lower for case A than case B: this means that overall, the specific heat capacity of object A must be larger than that of object B.
b) [tex]\frac{C_A}{C_B}=\frac{3.5(T_{oi}-T_{eq}+9)}{2.6(T_{oi}-T_{eq})}[/tex]
We can also give a quantitative comparison of the two specific heat capacities.
For object A we have:
[tex]C_A=\frac{3.5 m_w C_w}{m_o(T_{oi}-T_{eq})}[/tex]
For object B:
[tex]C_B=\frac{2.6 m_w C_w}{m_o(T_{oi}-T_{eq}')}[/tex]
Also, we know that the equilibrium temperature for object A is
[tex](3.5-2.6)=0.9^{\circ} C[/tex] higher than in case B, so we can write the second equation as
[tex]C_B=\frac{2.6 m_w C_w}{m_o(T_{oi}-(T_{eq}-0.9))}[/tex]
Now we can calculate the ratio of the two specific heat capacities:
[tex]\frac{C_A}{C_B}=\frac{3.5 m_w C_w}{m_o(T_{oi}-T_{eq})} \cdot \frac{m_o(T_{oi}-(T_{eq}-9))}{2.6 m_w C_w}=\frac{3.5(T_{oi}-T_{eq}+9)}{2.6(T_{oi}-T_{eq})}[/tex]
