An unknown metal, heated to an unknown temperature until a liquid is dropped
into a calorimeter that has 250 ml of water. The water heats up from 22 degrees
Celcius to 29 degrees Celcius, at which point the temperature of the water and
metal are equal. How much energy in Kilojoules did the metal release?
q = mass water x specific heat water x delta T water.
Note the correct spelling of celsius.
To calculate the amount of energy released by the metal, we need to use the equation:
Q = m * c * ΔT
Where:
Q = energy released by the metal (in kilojoules)
m = mass of water (in grams)
c = specific heat capacity of water (in joules/gram°C)
ΔT = change in temperature of the water (in °C)
First, let's convert the volume of water in the calorimeter (250 ml) to grams. We'll assume that the density of water is 1 gram/ml:
Mass of water (m) = volume of water * density of water
Mass of water (m) = 250 ml * 1 g/ml
Mass of water (m) = 250 g
Next, we need to calculate the change in temperature of the water:
ΔT = final temperature - initial temperature
ΔT = 29°C - 22°C
ΔT = 7°C
The specific heat capacity of water (c) is approximately 4.184 J/g°C.
Now, let's calculate the energy released by the metal:
Q = m * c * ΔT
Q = 250 g * 4.184 J/g°C * 7°C
Q = 73220 J
To convert the energy from joules to kilojoules, divide by 1000:
Q = 73220 J / 1000
Q = 73.22 kJ
Therefore, the metal released approximately 73.22 kilojoules of energy.