A 500 g piece of metal at 60 degrees celsius is placed in 200g of water at 22 degrees celsius contained in a coffee cup calorimeter. The metal and water come to the same temperature at 32.5 degrees cesius. The specific heat of the water is 4.184 j/g(degrees celsius)
a) how much heat energy did the metal give up to the water?
b) what is the specific heat capacity of the metal?
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[mass metal x specific heat metal x (Tfinal-Tinitial)] +[(mass water x specific heat water x (Tfinal-Tinitial)] = 0
Solve for specific heat metal, the only unknown in the equation.
The first set of brackets is the loss of heat from the metal, the second set of brackets is the gain of heat by water.
The metal and water come to the same temperature at 32.5 degrees cesius how do you do this part
To solve this problem, we can use the equation:
q = m * c * ΔT
where q is the heat energy transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
a) To find the heat energy transferred from the metal to the water, we can use the equation:
q = m * c * ΔT
Where m is the mass of the metal, c is the specific heat capacity of the water, and ΔT is the change in temperature.
Given:
Mass of metal (m) = 500 g
Specific heat capacity of water (c) = 4.184 J/g(°C)
Change in temperature (ΔT) = (final temperature - initial temperature) = (32.5°C - 60°C)
Substituting the values into the equation:
q = 500 g * 4.184 J/g(°C) * (32.5°C - 60°C)
q = 500 g * 4.184 J/g(°C) * (-27.5°C)
q = -57020 J
Therefore, the metal gave up -57020 J of heat energy to the water.
b) To find the specific heat capacity of the metal, we can use the equation:
q = m * c * ΔT
Given:
Heat energy transferred (q) = -57020 J
Mass of the metal (m) = 500 g
Change in temperature (ΔT) = (final temperature - initial temperature) = (32.5°C - 60°C)
Rearranging the formula to solve for c:
c = q / (m * ΔT)
Substituting the values into the equation:
c = -57020 J / (500 g * (32.5°C - 60°C))
c = -57020 J / (-500 g * 27.5°C)
c = 0.074 J/g(°C)
Therefore, the specific heat capacity of the metal is 0.074 J/g(°C).
To find the answers to both questions, we can use the principle of heat transfer, which states that the heat gained by one object is equal to the heat lost by another object in a closed system.
a) To determine how much heat energy the metal gave up to the water, we need to calculate the heat gained by the water. The formula for heat transfer, also known as the heat energy equation, is:
Q = mcΔT
Where:
Q = Heat energy gained or lost (in Joules)
m = Mass of the substance (in grams)
c = Specific heat capacity of the substance (in J/g°C)
ΔT = Change in temperature (final temperature - initial temperature) (in °C)
In this case, the water gained heat energy, so we can use the formula with the values provided:
m of water = 200g
c of water = 4.184 J/g°C
ΔT of water = (final temperature - initial temperature)
= (32.5°C - 22°C)
Now we can calculate the heat energy gained by the water:
Q_water = (200g) × (4.184 J/g°C) × (32.5°C - 22°C)
b) To determine the specific heat capacity of the metal, we use the same heat energy equation:
Q = mcΔT
This time, the heat energy is lost by the metal, so we can write the equation as:
Q_metal = (500g) × (c_metal) × (32.5°C - 60°C)
We have the value for Q_water from part (a). Since the heat energy lost by the metal is equal to the heat energy gained by the water (according to the principle of heat transfer), we can equate the two equations:
Q_metal = Q_water
From here, we can solve for the specific heat capacity of the metal, c_metal, by rearranging the equation:
(c_metal) = (Q_water) / [(500g) × (32.5°C - 60°C)]
Now, substitute the calculated values and solve for the specific heat capacity of the metal to get the answer to part (b).
I hope this explanation helps! If you have any further questions, please feel free to ask.