calculate the heat released for 5.00 grams if H2O cooling from 99 degrees celcius to 22 degrees celcius in joules and calories
Note the correct spelling of celsius.
q = mass x specific heat water x (Tfinal-Tinitial)
For calories specific heat = 1 cal/g*C
For joules specific heat = 4.184 J/g*C
How many calories are required to change 46 g of water from 7°C to 45°C?
To calculate the heat released during the cooling process, we can use the equation:
q = m * c * ΔT
where:
q = heat released (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
First, let's calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 22°C - 99°C
ΔT = -77°C
Now, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Next, plug in the values into the equation:
q = 5.00 g * 4.18 J/g°C * -77°C
q = -1596.70 J
The negative sign indicates that heat is being released, as the water is cooling down. The magnitude of the heat released is 1596.70 J.
To convert the heat from joules to calories, we can use the conversion factor:
1 calorie = 4.184 J
Let's convert:
q_calories = q_joules / conversion factor
q_calories = -1596.70 J / 4.184 J
q_calories ≈ -381.69 calories
So, the heat released is approximately 1596.70 joules (J) or -381.69 calories (cal).