The cube of iron and the cube of brick have the same dimensions. Each cube has sides of length l = 10 cm. What is the volume of each cube? The density of iron is ρFe= 7.87 g/cm3. Calculate the mass of the iron cube using the definition of density: ρ = m/V. Hence, mFe = 7.78+1000= 7830 gram The density of brick is ρbr= 1.92 g/cm3. Calculate the mass of the brick cube using the definition of density: ρ = m/V. Show your calculation. Hence, mFe = 1.92 X 1000 = 1920 gram
The cube of iron and the cube of brick have the same dimensions. Each cube has sides of length l = 10 cm. What is the volume of each cube?
- The density of iron is ρFe= 7.87 g/cm3. Calculate the mass of the iron cube using the definition of density: ρ = m/V.
Hence, mFe = 7.78+1000= 7830 gram
- The density of brick is ρbr= 1.92 g/cm3. Calculate the mass of the brick cube using the definition of density: ρ = m/V. Show your calculation.
Hence, mFe = 1.92 X 1000 = 1920 gram
- The water and the oil are both in beakers which have a 3-liter capacity. Each of the major tick marks on the beaker is 1 liter, and each minor tick mark is 0.2 liters. Examine the beakers and determine the volume (in ml) of each liquid.
1 mL= 1 cm^3 = 1cc
3 lt = 3x1000 mL= 1000 x 3cm^3= 3000 cc
Because the thermometer has no scale, we’ll have to calibrate the thermometer. Assume the experiment occurs at a (room) temperature of 77 oF.
- water reaches the boiling point, T= 100 °C. Notice that once the water reaches the boiling point, the temperature stops increasing. You can click on the Pause button so you can see the thermometer clearly.
- The bottom-most (first) tick mark on the thermometer corresponds to the initial room temperature, Ti and the nth tick mark corresponds to T = 100°C (i.e. Boiling point of water.) What is the value of each tick mark?
Answer: ∵ Thermometer is at 1st tick mark at room temp. = 77 °F = <?> °C, 100-27=73 °C
Total number of marks = 12
∆T = 100-27 °C corresponds to 12 tick marks. ∴ 1 tick mark = 6.08 °C.
- Calculate the amount of energy the cube of iron will absorb when raised from Tito T = 150 oC. [NB: The specific heat of iron is 0.113 cal/(g·oC). Use the formula below. Show your calculation.
QFe = mFecFeΔT = mFecFe(TFe – Ti ) ①
Answer: ∵ Ti = <?> °C and T(Fe) = 150 °C, ∴ we get: ∆T = (150 – <?>) °C = <?> °C. ∴ QFe = <?>
- Click on “Energy Symbols” and “Link Heaters”. Heat both cubes until the iron reaches TFe = 150°C. When the iron cube reaches TFe = 150°C, quickly release the heat slider and click on the Pause button.
- When the iron cube reaches TFe= 150°C, what is the temperature of the brick cube?
Answer: When TFe = 150 °C, TBrick = <?> °C (with the same amount of
heat input to both materials).
- Calculate the specific heat cbrfor the brick, using the same method used for the iron cube. Because the heaters were linked, the total energy absorbed by each cube is the same:
∴ Qbr = mbrcbrΔT = mbrcbr(Tbr − Ti) = QFe
∴ QFe = mbrcbr(Tbr − Ti)
∴ cbr = QFe/[mbr(Tbr − Ti)] ②
Use the values of mbr, QFe and Tbr from the previous calculations to determine the specific heat of the brick, cbr. Show your calculation.
Answer: Substituting the values into Eq’n ②, the specific heat of brick is
found to be: cBrick =
cBrick =
- Using the internet, find out the specific heat of the brick material.
Answer: cBrick = <?> ‡
‡ Cf: <enter URL as your citation here. ½ pt subtracted for no citation!>
- What is the percentage difference between the experimental and established values of the specific heat of brick? Show your calculation.
Answer: Percentage ∆ = × 100 = <?
- If the specific heat of brick is > the specific heat of iron, how is it possible for the brick cube to have a higher final temperature than the iron cube? Explain.
Procedure
2.1 i Repeat Experiment 1 with water and olive oil, to determine the specific heat coil of the olive oil.
Unlike iron, water can only heat up to T(H₂O) = 100 °C. The total amount of energy the water absorbs is:
∵ Q(H₂O)= m(H₂O) c(H₂O) ΔT = m(H₂O) c(H₂O) (T(H₂O) − Ti)
∴ Q(H₂O) = m(H₂O) [1 cal/(g⋅°C)] (100 °C − Ti)
- Use the mass of water m(H₂O)from Experiment 1 to calculate the amount of energy the water will absorb when raised from Ti (25 °C) to T(H₂O) = 100 °C.
Answer: Q(H₂O) = <?>
- Click on “Energy Symbols” and “Link Heaters”. Place the thermometers beside each beaker. Heat the oil and water beakers until the water reaches T(H₂O) = 100 °C. Release the heat slider and click the Pause button to measure the temperature of the oil, T(Oil).
- When the water reaches T(H₂O) = 100 °C, what is the temperature of the olive oil?
Answer: The temp. of the oil at that point is <?>
- Calculate the specific heat coilfor the olive oil. The total energy absorbed by the oil is the same as the energy absorbed by the water:
∵ Qoil = moil coil ΔT = moil coil (Toil − Ti) = QH₂O
∴ QH₂O = moil coil (Toil − Ti)
∴ coil = QH₂O÷[moil(Toil − Ti)] ④
Answer: Substituting the values into Eq’n ④, the specific heat of oil is
found to be: coil = <?>
coil = <?>
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