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?  

1. The density of iron is ρ_Fe= 7.87 g/cm³.  Calculate the mass of the iron cube using the definition of density: ρ = m/V.  

                  Hence, m_Fe = 7.78+1000= 7830 gram

1. The density of brick is ρ_br= 1.92 g/cm³.  Calculate the mass of the brick cube using the definition of density: ρ = m/V.  Show your calculation.

                  Hence, m_Fe = 1.92 X 1000 = 1920 gram

1. 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.

1. 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.
2. The bottom-most (first) tick mark on the thermometer corresponds to the initial room temperature, T_i and the n^th 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 1^st 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.

1. Calculate the amount of energy the cube of iron will absorb when raised from T_ito T = 150 ^oC.  [NB: The specific heat of iron is 0.113 cal/(g·^oC).  Use the formula below.  Show your calculation.

Q_Fe = m_Fec_FeΔT = m_Fec_Fe(T_Fe – T_i )                                   ①

Answer:   ∵  T_i = <?> °C and T_(Fe) = 150 °C,  ∴  we get:  ∆T = (150 – <?>) °C  = <?> °C.  ∴  Q_Fe = <?>

1. Click on “Energy Symbols” and “Link Heaters”. Heat both cubes until the iron reaches T_Fe = 150°C.  When the iron cube reaches T_Fe = 150°C, quickly release the heat slider and click on the Pause button. 
2. When the iron cube reaches T_Fe= 150°C, what is the temperature of the brick cube?

Answer:   When T_Fe = 150 °C, T_Brick = <?> °C (with the same amount of

                  heat input to both materials).

1. Calculate the specific heat c_brfor 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:

∴  Q_br = m_brc_brΔT = m_brc_br(T_br − T_i) = Q_Fe                                         
∴  Q_Fe = m_brc_br(T_br − T_i)                                                      
∴  c_br = Q_Fe/[m_br(T_br − T_i)]                                                ②

Use the values of m_br, Q_Fe and T_br from the previous calculations to determine the specific heat of the brick, c_br.  Show your calculation.

Answer:   Substituting the values into Eq’n ②, the specific heat of brick is

                  found to be:  c_Brick =

                                        c_Brick =

1. Using the internet, find out the specific heat of the brick material.

Answer:   c_Brick =    <?> ^‡ 

                  ^‡ Cf:  <enter URL as your citation here. ½ pt subtracted for no citation!> 

1. What is the percentage difference between the experimental and established values of the specific heat of brick? Show your calculation.

Answer:   Percentage ∆ =  × 100 = <?

1. 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 c_oil 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) − T_i) 
∴  Q_(H₂O) = m_(H₂O)[1 cal/(g⋅°C)] (100 °C − T_i)    

1. Use the mass of water m_(H₂O)from Experiment 1 to calculate the amount of energy the water will absorb when raised from T_i (25 °C) to T_(H₂O) = 100 °C.

Answer:   Q_(H₂O) = <?>

1. 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).
2. 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 <?>

1. Calculate the specific heat c_oilfor the olive oil. The total energy absorbed by the oil is the same as the energy absorbed by the water:

∵  Q_oil = m_oilc_oilΔT = m_oilc_oil(T_oil − T_i) = Q_H₂O
∴  Q_H₂O = m_oilc_oil(T_oil − T_i)
∴  c_oil = Q_H₂O÷[m_oil(T_oil − T_i)]                                             ④

Answer:   Substituting the values into Eq’n ④, the specific heat of oil is

                  found to be:  c_oil =  <?>

                                       c_oil =  <?>