Question

Two solid objects, A and B, are placed in boiling water and al-lowed to come to the temperature of the water. Each is then lifted out and placed in separate beakers containing 1000 g water at 10.0 °C. Object A increases the water temperature by 3.50 °C; B increases the water temperature by 2.60 °C. (a) Which object has the larger heat capacity? (b) What can you say about the specific heats of A and B?

Answer 1

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 $\Delta T$, according to the equation :

$Q=mC_s \Delta T$

where  $C_s$ is the specific heat capacity of the substance .

The equation can be rewritten as

$C_s = \frac{Q}{m \Delta T}$

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:

$m_w C_w (T_{eq} - T_{wi})=m_o C_o (T_{oi}-T_{eq})$

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

$m_o$ is the mass of the object

$C_o$ is the specific heat capacity of the object

$T_{oi}$ is the initial temperature of the object

Solving for $C_o$,

$C_o=\frac{m_w C_w (T_{eq} - T_{wi})}{m_o(T_{oi}-T_{eq})}$

For object A, the increase in temperature of the water is

$T_{eq}-T_{wi}=3.50^{\circ}$

So the formula becomes

$C_A=\frac{3.5 m_w C_w}{m_o(T_{oi}-T_{eq})}$

For object B, the increase in temperature of the water is

$T_{eq}-T_{wi}=2.60^{\circ}$

So the formula becomes

$C_B=\frac{2.6 m_w C_w}{m_o(T_{oi}-T_{eq})}$

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 $T_{oi}-T_{eq}$ 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) $\frac{C_A}{C_B}=\frac{3.5(T_{oi}-T_{eq}+9)}{2.6(T_{oi}-T_{eq})}$

We can also give a quantitative comparison of the two specific heat capacities.

For object A we have:

$C_A=\frac{3.5 m_w C_w}{m_o(T_{oi}-T_{eq})}$

For object B:

$C_B=\frac{2.6 m_w C_w}{m_o(T_{oi}-T_{eq}')}$

Also, we know that the equilibrium temperature for object A is

$(3.5-2.6)=0.9^{\circ} C$ higher than in case B, so we can write the second equation as

$C_B=\frac{2.6 m_w C_w}{m_o(T_{oi}-(T_{eq}-0.9))}$

Now we can calculate the ratio of the two specific heat capacities:

$\frac{C_A}{C_B}=\frac{3.5 m_w C_w}{m_o(T_{oi}-T_{eq})} \cdot \frac{m_o(T_{oi}-(T_{eq}-9))}{2.6 m_w C_w}=\frac{3.5(T_{oi}-T_{eq}+9)}{2.6(T_{oi}-T_{eq})}$