How much heat is added to 10 g of water when you start from room temperature and heat it to its boiling point?
We need to find the specific heat capacity of water. This can be found in google or textbooks.
specific heat of water = 4.184 J/g-K
Then we use the formula
Q = m*c*(T2 - T1)
where
Q = energy (J)
m = mass (g)
c = specific heat capacity (J/g-K)
T = temperature
I'm not sure about the room temperature. Some references say it's 25 deg C, while some say it's 20 deg C. But anyway, I'll just use 25 deg C.
The boiling point of water is 100 deg C.
Substituting,
Q = 10 * 4.184 * (100 - 25)
Q = ?
hope this helps~ `u`
To calculate the amount of heat added to water when heated from room temperature to its boiling point, we need to use the equation:
Q = mcΔT
Where:
Q represents the amount of heat added,
m is the mass of the substance (in this case, water),
c is the specific heat capacity of the substance (in this case, water),
and ΔT is the change in temperature.
Let's break it down step by step:
1. Determine the mass of the water: In this case, it is given as 10 g.
2. Find the specific heat capacity of water: The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius (J/g°C).
3. Calculate the change in temperature: The change in temperature is the difference between the final and initial temperatures. Since we are heating the water from room temperature (typically around 25°C) to its boiling point (100°C), the change in temperature would be (100°C - 25°C) = 75°C.
4. Substitute the values into the equation: Q = (10 g) * (4.18 J/g°C) * (75°C).
Calculating this further, we get:
Q = 3135 Joules
Therefore, approximately 3135 Joules of heat would be added to the 10 g of water when heated from room temperature to its boiling point.