Question

3. During an experiment, 98 g of water is used in the Styrofoam cup. The initial temperature

of water was 23.7°C. A 39.9-g piece of metal with initial temperature of 100.3°C (after
removing from the boiling water) is added to the calorimeter. The final temperature of water
was 28.2°C.

Calculate the specific heat of the metal.

Identify the metal.

Calculate the percent of error.

Answer 1

Step-by-step explanation:

To solve this problem, we can use the equation:

q = mcΔT

where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

First, we can calculate the heat absorbed by the water:

q_water = (98 g) (4.18 J/g°C) (28.2°C - 23.7°C) = 1938.4 J

Next, we can calculate the heat released by the metal:

q_metal = -q_water = -1938.4 J

The negative sign indicates that the metal released heat to the water.

We can rearrange the equation to solve for the specific heat of the metal:

c = -q_metal / (mΔT)

c = -(-1938.4 J) / (39.9 g) (100.3°C - 28.2°C)

c = 0.387 J/g°C

Therefore, the specific heat of the metal is 0.387 J/g°C.

To identify the metal, we can compare its specific heat to known values for different metals. From the specific heat values of common metals, we can see that the metal is most likely aluminum (specific heat of 0.902 J/g°C), as it has a much lower specific heat than copper, iron, or lead.

To calculate the percent error, we can use the formula:

percent error = (|experimental value - accepted value| / accepted value) x 100%

The accepted specific heat of aluminum is 0.902 J/g°C.

percent error = (|0.387 J/g°C - 0.902 J/g°C| / 0.902 J/g°C) x 100%

percent error = 57.1%

Therefore, the percent error is 57.1%.