Question

Suppose that 0.48 g of water at 25 ∘ C condenses on the surface of a 55- g block of aluminum that is initially at 25 ∘ C . If the heat released during condensation goes only toward heating the metal, what is the final temperature (in degrees Celsius) of the metal block? (The specific heat capacity of aluminum, C s,Al , is 0.903 J/(g⋅ ∘ C) .) Express the temperature in degrees Celsius to two significant figures.

Answer 1

Answer:

$49^oC$

Explanation:

At $25^oC$, the heat of vaporization of water is given by:

$\Delta H^o_{vap} = 43988 J/mol\cdot \frac{1 mol}{18.016 g} = 2441.6 J/g$

The water here condenses and gives off heat given by the product between its mass and the heat of vaporization:

$Q_1 = \Delta H^o_{vap} m_w$

The block of aluminum absorbs heat given by the product of its specific heat capacity, mass and the change in temperature:

$Q_2 = c_{Al}m_{Al}(t_f - t_i)$

According to the law of energy conservation, the heat lost is equal to the heat gained:

$Q_1 = Q_2$ or:

$\Delta H^o_{vap} m_w = c_{Al}m_{Al}(t_f - t_i)$

Rearrange for the final temperature:

$\Delta H^o_{vap} m_w = c_{Al}m_{Al}t_f - c_{Al}m_{Al}t_i$

We obtain:

$\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i = c_{Al}m_{Al}t_f$

Then:

$t_f = \frac{\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i}{c_{Al}m_{Al}} = \frac{2441.6 J/g\cdot 0.48 g + 0.903 \frac{J}{g^oC}\cdot 55 g\cdot 25^oC}{0.903 \frac{J}{g^oC}\cdot 55 g} = 49^oC$