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2 years ago

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The volume of a gas is decreased from 100 liters at 173.0°C to 50 liters at a constant pressure. After the decrease in volume, w

hat is the new temperature of the gas?

1 answer:

Minchanka [31]2 years ago

5 0

Answer:

223.08 K

Explanation:

First we <u>convert 173.0 °C to K</u>:

173.0 °C + 273.16 = 446.16 K

With the absolute temperature we can use <em>Charles' law</em> to solve this problem:

T₁V₂=T₂V₁

Where in this case:

T₁ = 446.16 K

V₂ = 50 L

T₂ = ?

V₁ = 100 L

We <u>input the data</u>:

446.16 K * 50 L = T₂ * 100 L

And <u>solve for T₂</u>:

T₂ = 223.08 K

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A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

OLEGan [10]

Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

Now, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$

Also,

mass of Ni, m = 15.0 g,

Initial temperature, $t_{i} = 100^{o}C$ ,

Final temperature, $t_{f}$ = ?

Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

$q_{water}(gain) = -q_{alloy}(lost)$

$55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$ = -( $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$ )

$t_{f} = \frac{25.9^{o}C}{1.029}$

= $25.2^{o}C$

Thus, we can conclude that the final temperature of nickel and water is $25.2^{o}C$ .

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