Which of the following would not be considered an observation in terms of the scientific method?
1 answer:
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Answer:
The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g
Explanation:
The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid
The latent heat of evaporation of water, Δ ≈ 2,230 J/g
The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows
Q = m × Δ
Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;
6,768 J = m × 2,230 J/g
∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g
The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.
<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C
<u>Explanation:</u>
When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.
The equation used to calculate heat released or absorbed follows:
......(1)
where,
q = heat absorbed or released
= mass of hot water = 46.7 g
= mass of cold water = 45.33 g
= final temperature = 59.4°C
= initial temperature of hot water = 80.6°C
= initial temperature of cold water = 40.6°C
= specific heat of hot water = 4.184 J/g°C
= specific heat of cold water = 4.184 J/g°C
= specific heat of calorimeter = ? J/g°C
Putting values in equation 1, we get:
Hence, the specific heat of calorimeter is 30.68 J/g°C