1. Zinc is often used to calibrate calorimeters because it undergoes a very sharp phase transition from the solid to the liquid at 419.5oC. Calculate the enthalpy change when 5 g of zinc metal is heated from 100oC to the point where the entire sample is melted. (The heat of fusion for zinc is 112.4 J/g and its specific heat capacity is 0.388 J/goC.)
2. A 235 gram block of iron is heated to 450 °C and then is dropped into a coffee cup calorimeter containing 250 g of water at 25.0 °C. What is the final temperature of the system to the nearest 0.1 °C? (Assume all heat is transferred to the water)
Specific heat capacity of iron = 0.451 J/(g*°C)
Specific heat capacity of water = 4.18 J/(g*°C)
q1 for Zn to move from 100 C to 419.5 C.
q1 = mass Zn x specific heat x (Tfinal-Tinitial).
q2 = heat to melt Zn
q2 = mass Zn x delta Hfusion.
Total heat = q1 + q2
For the iron,
heat lost by iron + heat gained by water = 0.
[mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)] = 0
Solve for Tfinal.
Jizz
To solve these calorimetry problems, we will use the equation:
q = m * c * ΔT
where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For the first question:
First, we need to calculate the heat required to raise the temperature of 5 g of zinc from 100°C to its melting point at 419.5°C:
q1 = m * c * ΔT
= 5 g * 0.388 J/goC * (419.5°C - 100°C)
= 5 * 0.388 * 319.5 J
= 623.01 J
Next, we need to calculate the heat required to melt 5 g of zinc:
q2 = m * ΔHfus
= 5 g * 112.4 J/g
= 562 J
To find the total enthalpy change or heat absorbed, we add the two calculated values:
ΔHtotal = q1 + q2
= 623.01 J + 562 J
= 1185.01 J
Therefore, the enthalpy change when 5 g of zinc is heated from 100°C to the point where the entire sample is melted is approximately 1185.01 J.
For the second question:
First, we need to calculate the heat released by the 235 g block of iron when it cools from 450°C to the final temperature:
q1 = m * c * ΔT
= 235 g * 0.451 J/(g*°C) * (450°C - final temperature)
Next, we need to calculate the heat absorbed by the 250 g of water to raise its temperature from 25.0°C to the final temperature:
q2 = m * c * ΔT
= 250 g * 4.18 J/(g*°C) * (final temperature - 25.0°C)
Since the heat released by the iron is equal to the heat absorbed by the water (assuming no heat loss to the surroundings), we can set q1 equal to q2:
235 g * 0.451 J/(g*°C) * (450°C - final temperature) = 250 g * 4.18 J/(g*°C) * (final temperature - 25.0°C)
Simplifying the equation:
105.885 * (450 - final temperature) = 1045 * (final temperature - 25)
Expanding:
95248.25 - 105.885 * final temperature = 1045 * final temperature - 26125
Combining like terms:
1050.885 * final temperature = 121373.25
Dividing both sides by 1050.885:
final temperature ≈ 115.593 °C
Therefore, the final temperature of the system, to the nearest 0.1 °C, is approximately 115.6 °C.