A small pebble is heated and placed in a foam cup calorimeter containing 25.0 mL of water at 25.0 degrees Celsius.The water reaches a maximum temperature of 26.4 degrees Celsius. How many joules of heat were released by the pebble?
146.3
q = mass water x specific heat water x delta T.
You know mass, specific heat water is 4.184 J/g*C and delta T is Tfinal-Tinitial.
Volume of Water: 25 mL= 25 g
T1= 25.0 C
T2= 26.4
ΔT= 1.4 C
Q= m CΔT
=25 g x Δ1.8 (26.4-25.0)
=25 x Δ1.8 x1.4= 146.3
Well, well, well, looks like we have a hot pebble here! Let's do some math and find out how much heat it released.
First of all, we need to calculate the change in temperature (delta T) of the water. So, 26.4 degrees Celsius minus 25.0 degrees Celsius gives us a delta T of 1.4 degrees Celsius.
Since we're dealing with water, we know that its specific heat capacity is 4.18 J/g°C. And we also know that the water in our foam cup calorimeter has a volume of 25.0 mL, which means it has a mass of 25.0 g. Voila!
Now, to calculate the heat released, we use the equation Q = m x c x delta T, where Q is the heat released, m is the mass, c is the specific heat capacity, and delta T is the change in temperature.
So, Q = 25.0 g x 4.18 J/g°C x 1.4°C. Go ahead, do the math.
Okay, the calculation gives us 146.5 J (rounded to one decimal place). This means that our little pebble released approximately 146.5 joules of heat. That's quite a spicy pebble, huh?
To find the amount of heat released by the pebble, you can use the equation:
q = m * c * ΔT
where:
q is the amount of heat released (in joules),
m is the mass of water (in grams),
c is the specific heat capacity of water (which is 4.18 J/g°C), and
ΔT is the change in temperature (in degrees Celsius).
In this case, the mass of water is equal to its volume multiplied by its density. The density of water is approximately 1 g/mL.
First, calculate the mass of the water:
mass = volume * density
mass = 25.0 mL * 1 g/mL
mass = 25.0 g
Next, calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 26.4°C - 25.0°C
ΔT = 1.4°C
Now, substitute the values into the equation:
q = m * c * ΔT
q = 25.0 g * 4.18 J/g°C * 1.4°C
q = 146.5 J
Therefore, the pebble released approximately 146.5 joules of heat.