A 55.6g sample of water is cooled from 50 degrees C to 35.5 degrees C.
How many calories released?
How many kilocalories?
How many joules?
How many calories are required to raise the temperature of 205 grams of water from 35.5 degrees celsius to 80.0 degrees celsius?
To determine the amount of heat released during the cooling process, we can use the formula:
q = m * c * ΔT
where:
q is the heat released,
m is the mass of the water sample (55.6g),
c is the specific heat capacity of water (1 calorie/gram °C), and
ΔT is the change in temperature (50 °C - 35.5 °C).
Let's calculate the answers step by step:
1. Calories released:
q = m * c * ΔT
q = 55.6g * 1 cal/g°C * (50°C - 35.5°C)
q = 55.6g * 1 cal/g°C * 14.5°C
q = 806.2 calories
Therefore, 806.2 calories are released during the cooling process.
2. Kilocalories released:
To convert calories to kilocalories, we divide the answer from step 1 by 1000.
Kilocalories = 806.2 calories / 1000
Kilocalories = 0.8062 kilocalories
Therefore, 0.8062 kilocalories are released during the cooling process.
3. Joules released:
To convert calories to joules, we use the conversion factor: 1 calorie = 4.184 joules.
Joules = 806.2 calories * 4.184 joules/calorie
Joules = 3,371.0968 joules
Therefore, 3,371.0968 joules are released during the cooling process.
To summarize:
- The water sample releases 806.2 calories during the cooling process.
- This is equivalent to 0.8062 kilocalories.
- This is equivalent to 3,371.0968 joules.