If the warm water in this experiment is at 98 °C, how much heat must be released in order for the temperature of 100. mL of water to decrease to 32°C?
Well, let's see... to heat up water from 98°C to 100°C and then cool it down to 32°C, it's like a roller coaster ride for water molecules! They go up and they go down! So, the amount of heat released will depend on the specific heat capacity of water.
But fear not! Clown Bot is here to crunch some numbers for you. The specific heat capacity of water is about 4.184 J/g°C. Since you have 100. mL of water (which is the same as 100 grams), we can use that to calculate the heat released.
The temperature change is 98°C - 32°C = 66°C. So, the amount of heat released can be determined using the formula:
Q = (mass) x (specific heat capacity) x (temperature change)
Q = (100 g) x (4.184 J/g°C) x (66°C)
Calculating that, we get:
Q = 277,584 J (rounded to the nearest whole number)
So, the amount of heat that must be released in order to decrease the temperature of 100 mL of water from 98°C to 32°C is approximately 277,584 joules. Lots of ups and downs for those water molecules!
To calculate the amount of heat that must be released for the temperature to decrease from 98 °C to 32 °C, we can use the formula:
Q = m * C * ΔT
Where:
Q = Heat energy
m = mass of water
C = specific heat capacity of water
ΔT = change in temperature
First, let's calculate the mass of water:
Given: Volume of water = 100 mL
1 mL of water is approximately equal to 1 gram (g). Therefore,
mass of water = 100 mL = 100 g
Next, let's find the change in temperature:
Initial temperature = 98 °C
Final temperature = 32 °C
ΔT = Final temperature - Initial temperature
ΔT = 32 °C - 98 °C
ΔT = -66 °C
Since the temperature is decreasing, the change in temperature is negative.
The specific heat capacity of water is approximately 4.18 J/(g°C).
Now, we can calculate the amount of heat released:
Q = m * C * ΔT
Q = 100 g * 4.18 J/(g°C) * (-66 °C)
Calculating this, we get:
Q = -275,880 J
So, approximately -275,880 Joules of heat must be released for the temperature of 100. mL of water to decrease from 98 °C to 32 °C.
To calculate the amount of heat that needs to be released in order for the temperature of the water to decrease, we can use the formula:
Q = mcΔT
where:
Q is the amount of heat energy absorbed or released (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in joules per gram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
In this case, we are given:
m = 100. mL of water (which is equivalent to 100. grams since 1 mL of water has a mass of 1 gram),
c = the specific heat capacity of water (which is approximately 4.18 J/g°C),
ΔT = final temperature (32°C) - initial temperature (98°C) = -66°C.
Plug the values into the formula:
Q = (100. g) x (4.18 J/g°C) x (-66°C)
Q = -275,928 joules
Therefore, approximately -275,928 joules of heat must be released in order for the temperature of 100. mL of water to decrease from 98°C to 32°C. The negative sign indicates that heat is being released.