Answer:
P = 20.1697 atm
Explanation:
In this case we need to use the ideal gas equation which is:
PV = nRT (1)
Where:
P: Pressure (atm)
V: Volume (L)
n: moles
R: universal gas constant (=0.082 L atm / K mol)
T: Temperature
From here, we can solve for pressure:
P = nRT/V (2)
According to the given data, we have the temperature (T = 20 °C, transformed in Kelvin is 293 K), the moles (n = 125 moles), and we just need the volume. But the volume can be calculated using the data of the cylinder dimensions.
The volume for any cylinder would be:
V = πr²h (3)
Replacing the data here, we can solve for the volume:
V = π * (17)² * 164
V = 148,898.93 cm³
This volume converted in Liters would be:
V = 148,898.93 mL * 1 L / 1000 mL
V = 148.899 L
Now we can solve for pressure:
P = 125 * 0.082 * 293 / 148.899
<h2>
P = 20.1697 atm</h2>
Answer:
Heat lost to the surroundings
Heat lost to the thermometer
Explanation:
All changes in heat, or energy, can be explained. Many of the reactions or changes we see in the world involve the conversion of energy. For example as we heat up a substance (eg. water), the amount of energy we put in should give us an exact temperature. However, this is a "perfect world" scenario, and does not occur in real life. Whenever heat is added to a substance like water, we always need to account for the energy that is going to be lost. For example, heat lost to evaporation or even the effect of measuring the temperature with a thermometer (the introduction of anything including a thermometer will affect the temperature).
Convert 57.6 L to dm3 and divide it by 24
<u>Answer:</u>
Specific heat of a substance is the value that describe how the added heat energy of substance has the impact on its temperature.
Unit is <em>
</em>
<em>C = Q/m. ∆T</em>
<em>C – Specific heat
</em>
<em>Q- heat energy (J)</em>
<em>M – Mass (Kg)</em>
<em>∆T- change in temperature (K) </em>
<u>Explanation:</u>
<em>Given data:</em>
<em>M= 140 g = 0.14 Kg</em>
<em>Q – 1080 Joules.</em>
<em>∆T – 98.4 – 62.2 = 36.2</em>
Substituting the given data in Equation
<em>Specific heat of Aluminium =
</em>