Which statement BEST describes what happens when a cold front approaches a warm front?

Chemistry

2 answers:

stepladder [879]3 years ago

7 0

Answer:

Explanation:

In a cold front set-up, the boundary between the cold and warm air masses is relatively steep, typically causing the warm air in front of it to rise rapidly.

katen-ka-za [31]3 years ago

5 0

Answer:

Explanation:

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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 th

mamaluj [8]

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:

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