W=0.40
X=0.60
Y=0.24
Z=0.90
6. A 100.0 gram sample of an unknown metal at a temperature of 75°C is dropped into a polystyrene cup containing 100.0 grams of water that was at an initial temperature of 22°C. After the addition of the metal sample and stirring, the temperature of the water plus metal reached a high of 25°C. Assume that the specific heat for water is 4.0 J/g°C. Referring to the chart above, which metal was used?
(A) W
(B) X
(C) Y
(D) Z
We can use the formula for heat transfer, which is given by:
q = mcΔT
where q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
For the water, we can calculate the heat transferred as:
q_water = (100.0g)(4.0 J/g°C)(25°C - 22°C) = 120 J
For the metal, we can calculate the heat transferred as:
q_metal = (100.0g)(c_metal)(25°C - 75°C)
Since the initial temperature of the water is lower than the final temperature, we can assume that the water lost heat and the metal gained heat. Therefore, the heat transferred to the metal should be equal to the heat transferred from the water, i.e., q_metal = -q_water.
We can solve for c_metal as follows:
(100.0g)(c_metal)(-50°C) = 120 J
c_metal = -120 J / (100.0g * -50°C)
c_metal = 0.24 J/g°C
Comparing the value of c_metal to the given values from the chart, we can see that it matches with the value of Y (0.24).
Therefore, the metal used is (C) Y.
To determine which metal was used, we can calculate the amount of heat transferred during the process using the formula:
q = m * c * ΔT
where:
q = heat transferred
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
First, let's calculate the amount of heat transferred to the water. The mass of the water is 100.0 grams, the specific heat capacity of water is 4.0 J/g°C, and the change in temperature is (25°C - 22°C) = 3°C.
q_water = 100.0 g * 4.0 J/g°C * 3°C
q_water = 1200 J
Next, let's calculate the amount of heat transferred to the metal. The mass of the metal is given as 100.0 grams. We assume that the metal has the same specific heat capacity as water (4.0 J/g°C) to simplify the calculation. The change in temperature for the metal is (25°C - 75°C) = -50°C.
q_metal = 100.0 g * 4.0 J/g°C * -50°C
q_metal = -20,000 J
Since the heat transferred to the metal is negative, it indicates that heat was lost by the metal.
To determine which metal was used, we compare the amount of heat transferred to the water for each option (W, X, Y, Z) with the amount of heat transferred to the metal (which is negative at -20,000 J).
For the options given:
(A) W:
q_water = 100.0 g * 4.0 J/g°C * (25°C - 22°C)
q_water = 1200 J
This does not match the heat transferred to the metal.
(B) X:
q_water = 100.0 g * 4.0 J/g°C * (25°C - 22°C)
q_water = 1200 J
This does not match the heat transferred to the metal.
(C) Y:
q_water = 100.0 g * 4.0 J/g°C * (25°C - 22°C)
q_water = 1200 J
This does not match the heat transferred to the metal.
(D) Z:
q_water = 100.0 g * 4.0 J/g°C * (25°C - 22°C)
q_water = 1200 J
This matches the heat transferred to the metal (-20,000 J).
Therefore, the metal used in the experiment is option (D) Z.