How many kilocalories of heat are released when 125 g of water cools from 100.0oC to 19.5oC?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Well, let me do some calculations while preparing my clown nose!
To calculate the amount of heat released, we can use the equation:
Q = mcΔT
Where Q is the amount of heat released, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is approximately 4.18 J/g°C.
First, let's convert the mass of water from grams to kilograms:
125 g = 0.125 kg
Now, let's calculate the amount of heat released:
Q = (0.125 kg) * (4.18 J/g°C) * (100.0 - 19.5 °C)
Calculating this, we get:
Q ≈ 450 Joules
But the question is asking for the answer in kilocalories. And guess what? 1 kilocalorie is equal to 4184 Joules!
So converting the answer, we have:
Q ≈ 450 Joules ÷ 4184 J/kcal ≈ 0.11 kilocalories
There you have it! Approximately 0.11 kilocalories of heat are released when 125 g of water cools from 100.0°C to 19.5°C.
To calculate the heat released when the water cools, we can use the formula:
Q = m * c * ΔT
Where:
Q = heat released (in calories or kilocalories)
m = mass of the water (in grams)
c = specific heat capacity of water (1 cal/g°C or 1 kcal/kg°C)
ΔT = change in temperature (in °C)
Given:
m = 125 g
c = 1 kcal/kg°C (since we want the answer in kilocalories)
ΔT = 100.0°C - 19.5°C = 80.5°C
Substituting these values into the formula:
Q = 125 g * 1 kcal/kg°C * 80.5°C
Q = 10,062.5 kcal
Therefore, when 125 g of water cools from 100.0°C to 19.5°C, approximately 10,062.5 kilocalories of heat are released.
To calculate the amount of heat released when a substance cools down, we can use the formula:
Q = mcΔT
Where:
Q is the heat released or absorbed,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
For water, the specific heat capacity is 4.18 J/g·°C. However, since the specific heat capacity is given in joules and the mass is given in grams, we need to convert the mass of water from grams to kilograms.
Given:
Mass of water (m) = 125 g = 0.125 kg
Initial temperature (T initial) = 100.0 °C
Final temperature (T final) = 19.5 °C
First, let's calculate the change in temperature:
ΔT = T final - T initial
ΔT = 19.5 °C - 100.0 °C
ΔT = -80.5 °C (Note: We take the negative sign because the water is cooling down)
Now, we can substitute the values into the formula and calculate the heat released (Q):
Q = mcΔT
Q = (0.125 kg) * (4.18 J/g·°C) * (-80.5 °C)
To convert J (joules) to kcal (kilocalories), we use the conversion: 1 kcal = 4184 J.
Q = (0.125 kg) * (4.18 J/g·°C) * (-80.5 °C) / 4184 J
Q ≈ -0.19 kcal
Therefore, when 125 g of water cools from 100.0 °C to 19.5 °C, approximately -0.19 kilocalories of heat are released.
(Note: The negative sign indicates that heat is released, not absorbed.)