An unknown piece of metal weighing 100. g is heated to 90.0 degrees Celcius. It is dropped into 250 g of water at 20.0 degrees Celcius. When equilibrium is reached, the temperature of both the water and piece of metal is 29.0 degrees Celcius. Determine the specific heat of the metal using the fact that the energy lost by the metal must equal the energy absorbed by the water. Assume that the heat capacity of the container, a Styrofoam cup, is negligible.
My answer: 1.543 J/g C
Use ΣmcΔT=0.
For metal,
mcΔT=100*c*(29-90)=6100c
For water,
mcΔT=250*4.184*(29-20)=9414
Applying ΣΔT=0
=>
100*c*(29-90)=250*4.184*(29-20)
Solving for c gives 1.543 J-g-1-°C-1 as you had it.
Correction: the answer is more appropriately given as 1.54 J/(g-°C) as all the original information is given to 3 significant figures.
To determine the specific heat of the metal, we can use the equation:
q metal = q water
where q is the heat transferred. The heat transferred can be calculated using the equation:
q = m * c * ∆T
where:
q is the heat transferred
m is the mass
c is the specific heat
∆T is the change in temperature
Let's start by calculating the heat transferred by the metal:
q metal = m metal * c metal * ∆T metal
Given:
m metal = 100. g (mass of the metal)
∆T metal = 29.0 - 90.0 = -61.0 degrees Celcius (change in temperature of the metal)
Next, let's calculate the heat transferred by the water:
q water = m water * c water * ∆T water
Given:
m water = 250. g (mass of the water)
c water = 4.18 J/g°C (specific heat of water)
∆T water = 29.0 - 20.0 = 9.0 degrees Celcius (change in temperature of the water)
Since the heat lost by the metal is equal to the heat gained by the water, we have:
q metal = q water
m metal * c metal * ∆T metal = m water * c water * ∆T water
Substituting the known values:
100. g * c metal * (-61.0°C) = 250. g * 4.18 J/g°C * 9.0°C
Simplifying:
-61.0 * c metal = 250 * 4.18 * 9.0
-61.0 * c metal = 9411
Dividing both sides by -61.0:
c metal ≈ 154.1 J/g°C
Therefore, the specific heat of the metal is approximately 154.1 J/g°C.
To determine the specific heat of the metal, we need to equate the energy lost by the metal to the energy gained by the water.
First, let's calculate the energy lost by the metal. We can use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy
m is the mass of the metal
c is the specific heat of the metal
ΔT is the change in temperature of the metal
Since the unknown metal piece is heated from 90.0°C to 29.0°C, the ΔT is:
ΔT = 29.0°C - 90.0°C = -61.0°C
Now let's calculate the energy lost by the metal:
Q_lost = m * c * ΔT
Substituting the known values:
Q_lost = 100.0 g * c * (-61.0°C)
Next, we need to determine the energy gained by the water. Again, we can use the same formula:
Q = m * c * ΔT
Where:
Q is the heat energy
m is the mass of the water (250.0 g)
c is the specific heat of water (4.184 J/g°C) -- Note: This value is known and can be looked up
ΔT is the change in temperature of the water
The water is heated from 20.0°C to 29.0°C, so:
ΔT = 29.0°C - 20.0°C = 9.0°C
Now let's calculate the energy gained by the water:
Q_gained = m * c * ΔT
Q_gained = 250.0 g * 4.184 J/g°C * 9.0°C
Now, equating the energy lost by the metal to the energy gained by the water, we have:
Q_lost = Q_gained
100.0 g * c * (-61.0°C) = 250.0 g * 4.184 J/g°C * 9.0°C
Simplifying the equation:
-61.0°C * c = 9.0°C * 250.0 g * 4.184 J/g°C
Solving for c:
c = (9.0°C * 250.0 g * 4.184 J/g°C) / (-61.0°C)
Evaluating the expression:
c ≈ 1.543 J/g°C
Therefore, the specific heat of the metal is approximately 1.543 J/g°C.