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3 years ago

If you have a 1500 g aluminum pot, how much heat energy is needed to raise its temperature by 100°C?

Chemistry

1 answer:

Nataly [62]3 years ago

8 0

The heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ.

The heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

Explanation:

The heat energy required to raise the temperature of any body can be obtained from the specific heat formula. As this formula states that the heat energy required to raise the temperature of the body is directly proportional to the product of mass of the body, specific heat capacity of the material and temperature change experienced by the material.

So in this problem, the mass of the aluminium is given as m = 1500 g, the specific heat of the aluminium is 0.900 J/g °C. Then as it is stated that the temperature is raised by 100 °C, so the pots are heat to increase by 100 °C from its initial temperature. This means the difference in temperature will be 100°C (ΔT = 100°C).

Then, the heat energy required to raise the temperature will be

$q = m*c*del T = 1500 * 0.900 * 100 = 135000 = 135 kJ$

Thus, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100 °C is 135 kJ.

Similarly, the mass of copper pot is given as 1500 g, the specific heat capacity of copper is 0.385 and the difference in temperature is 100 °C.

Then, the heat energy required to raise its temperature will be

$q = m*c*del T = 1500 * 0.385 * 100 = 57750 = 57.75 kJ$

And the heat energy required to raise the temperature of 1500 g of copper pot by 100°C is 57.75 kJ.

So, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ. And the heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

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Ilia_Sergeevich [38]

27) Partial pressure of oxygen: 57.8 kPa

29) Final volume: 80 mL

30) Final volume: 8987 L

31) Due to property of water of being polar, ice floats on water

Explanation:

27)

In a mixture of gases, the total pressure of the mixture is the sum of the partial pressures:

$p_T = p_1 + p_2 + ... + p_N$

In this problem, the mixture contains 3 gases (helium, carbon dioxide and oxygen). We know that the total pressure is

$p_T=201.4 kPa$

We also know the partial pressures of helium and carbon dioxide:

$P_{He}=125.4 kPa\\P_{CO_2}=18.2 kPa$

The total pressure can be written as

$p_T=p_{He}+p_{CO_2}+p_{O_2}$

where $p_{O_2}$ is the partial pressure of oxygen. Therefore, we find

$p_{O_2}=p_T-p_{He}-p_{CO_2}=201.4-125.4-18.2=57.8 kPa$

29)

Assuming that the pressure of the gas is constant, we can apply Charle's law, which states that:

"For an ideal gas at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically,

$\frac{V}{T}=const.$

where

V is the volume of the gas

T is the Kelvin temperature

We can re-write it as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Here we have:

$V_1 = 42 mL$ (initial volume)

$T_1=-89^{\circ}C+273=184 K$ is the initial temperature

$T_2=77^{\circ}C+273=350 K$ is the final temperature

Solving for V2, we find the final volume:

$V_2=\frac{V_1 T_2}{T_1}=\frac{(42)(350)}{184}=80 mL$

30)

For this problem, we can use the equation of state for ideal gases, which can be written as

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where in this problem:

$p_1 = 102.3 kPa$ is the initial pressure

$V_1=1975 L$ is the initial volume

$T_1=25^{\circ}C+273=298 K$ is the initial temperature

$p_2=21.5 kPa$ is the final pressure

$T_2=12^{\circ}C+273=285 K$ is the final temperature

And solving for V2, we find the final volume of the balloon:

$V_2=\frac{p_1 V_1 T_2}{p_2 T_1}=\frac{(102.3)(1975)(285)}{(21.5)(298)}=8987 L$

31)

A molecule of water consists of two atoms hydrogen bond with an atom of oxygen ( $H_2 O$ ) in a covalent bond.

While the molecul of water is overall neutral, due to the higher electronegativity of the oxygen atom, electrons are slightly shifted towards the oxygen atom; as a result, there is a slightly positive charge on the hydrogen side, and a slightly negative charge on the oxygen side (so, the molecules is said to be polar).

As a consequence, molecules of water attract each other, forming the so-called "hydrogen bonds".

One direct consequence of the polarity of water is that ice floats on liquid water.

Normally, for every substance on Earth, the solid state is more dense than the liquid state. However, this is not true for water, because ice is less dense than liquid water.

This is due to the polarity of water. In fact, when the temperature of water is decreased to freezing point and water becomes ice, the hydrogen bondings "force" the molecules to arrange in a lattice structure, so that the molecules become more spaced when they turn into solid state. As a result, ice occupies more volume than water, and therefore it is less dense, being able to float on water.

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Explanation:

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