near what like open flames? if so false unless it clarifies the solution requires alcohol then its a no go
The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C
<h3>How to calculate temperature?</h3>
The initial temperature of the copper metal can be calculated using the following formula on calorimetry:
Q = mc∆T
mc∆T (water) = - mc∆T (metal)
Where;
- m = mass
- c = specific heat capacity
- ∆T = change in temperature
According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:
400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)
30,096 = 93.6T - 2246.4
93.6T = 32342.4
T = 345.5°C
Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.
Learn more about temperature at: brainly.com/question/15267055
Answer:
The volume of the gas is determined, which will allow you to calculate the temperature.
Explanation:
According to Charles law; the volume of a given mass of an ideal gas is directly proportional to its temperature at constant pressure.
This implies that, when the volume of an ideal gas is measured at constant pressure, the temperature of the ideal gas can be calculated from it according to Charles law.
Hence in the Ideal Gas Law lab, the temperature of an ideal gas is measured by determining the volume of the ideal gas.
Answer:
140 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 3 atm
- Initial temperature of the gas (T₁): 280 K
- Final pressure of the gas (P₂): 1.5 atm
- Final temperature of the gas (T₂): ?
Step 2: Calculate the final temperature of the gas
We have a gas whose pressure is reduced. If we assume an ideal behavior, we can calculate the final temperature of the gas using Gay-Lussac's law.
T₁/P₁ = T₂/P₂
T₂ = T₁ × P₂/P₁
T₂ = 280 K × 1.5 atm/3 atm = 140 K
I think it’s water evaporation I’m not sure tho
