Heat Transfer Quiz - Answer Key

Complete Solutions with Work Shown

Section A Conceptual Understanding - Answers

Problem 1

Justify the reasoning behind the assumption that a metal placed in boiling water will eventually attain the same temperature as the boiling water.

The flow of heat always goes from higher temperature to lower temperature. In this scenario, the higher temperature is the boiling water which transfers heat to the lower temperature metal. This happens until the system reaches thermal equilibrium (both at the same temperature).
Key Points:
  • Heat naturally flows from hot → cold
  • Boiling water (100°C) → metal (initially cooler)
  • Process continues until thermal equilibrium is reached
  • At equilibrium, both substances are at 100°C (the boiling point of water)

Problem 2

Outline why it is important to transfer the heated metal to the calorimeter as quickly as possible.

It is important to transfer the heated metal to the calorimeter as quickly as possible because the second the metal leaves the boiling water, air and the calorimeter come into contact with the metal and it begins to lose heat. Quick transfer ensures the temperature is as accurate as possible and minimizes heat loss to the surroundings.
Detailed Explanation:
  • The metal begins cooling immediately upon removal from boiling water
  • Heat is lost to the surrounding air during transfer
  • Slower transfer → more heat lost → lower initial temperature in calorimeter
  • This leads to inaccurate specific heat calculations
  • Quick transfer minimizes this error and maintains experimental accuracy

Problem 3

Describe how the volume of water could be used to determine its mass.

The density of water is always 1.0 g/mL, which will be the same as the volume. Therefore, the mass of water in grams equals the volume in mL.
Calculation Method:
Using the density formula: \(D = \frac{m}{V}\)
Rearrange to: \(m = D \times V\)

Since \(D_{\text{water}} = 1.0 \text{ g/mL}\):
\(m = 1.0 \text{ g/mL} \times V_{\text{mL}}\)

Example: 150 mL of water = 150 g × 1.0 g/mL = 150 g

Problem 4

The method states that the thermometer should be checked regularly until thermal equilibrium has been reached. Suggest how an experimenter will know when thermal equilibrium has been attained and state the necessary measurement that is obtained from it.

Thermal equilibrium is reached when the temperature stops changing (remains constant). The experimenter monitors the thermometer and when the temperature stabilizes for a period of time, equilibrium has been reached. The necessary measurement is the final temperature of the system.
Practical Method:
  • Continuously monitor the thermometer after adding the hot metal
  • Temperature will initially rise quickly, then slow down
  • When temperature remains constant for 30-60 seconds, equilibrium is reached
  • Record this as \(T_{\text{final}}\) or \(T_{\text{equilibrium}}\)
  • This temperature is the same for both the metal and water

Problem 5

Explain the reasoning for changing the calorimeter water before each trial.

Fresh water should be used for each trial to ensure consistent starting conditions. The water from the previous trial has already been heated and would have a higher initial temperature, which would affect the accuracy of the experiment.
Additional Reasons:
  • Each trial needs the same initial water temperature for valid comparisons
  • Used water may have evaporated slightly, changing the mass
  • Fresh water ensures controlled variables across all trials
  • Allows for accurate determination of \(\Delta T\) for the water

Problem 6

Outline three realistic solutions that could minimize heat losses in this experiment.

  1. Use an insulated calorimeter (styrofoam cup or insulated container) to reduce heat loss to surroundings
  2. Transfer the metal as quickly as possible from boiling water to calorimeter
  3. Use a lid on the calorimeter to prevent heat loss through evaporation and convection
Additional Options:
  • Perform experiment in a draft-free environment
  • Use tongs or transfer device that minimize contact with the metal
  • Pre-warm the calorimeter slightly to reduce temperature gradient
  • Stir gently but consistently to ensure even temperature distribution

Section B Calculations Using the Data Table - Answers

Problem 7

Show, using the data from each trial, the determination of volume and density of the metal.

Volume is given directly in the data table. Density = mass ÷ volume
Trial 1 (detailed work):
Mass = 282.25 g
Volume = 20 mL = 20 cm³
\[D = \frac{m}{V} = \frac{282.25 \text{ g}}{20 \text{ cm}^3} = 14.1 \text{ g/cm}^3\]

Summary of All Trials:
Trial Mass (g) Volume (cm³) Density (g/cm³)
1 282.25 20 14.1
2 121.94 4 30.5
3 107.47 3 35.8
4 77.88 2 38.9
5 146.89 1 146.9

Note: Trial 5 has an unrealistic density, suggesting a measurement error in volume.

Problem 8

Show, using data from each trial, the calculation for the specific heat of the metal.

Trial 1 (detailed work):
Given Data:
Metal: m = 282.25 g, \(T_i = 100°C\), \(T_f = 22°C\)
Water: V = 170 mL → m = 170 g, \(c = 4.18\) J/g°C, \(T_i = 20°C\), \(T_f = 22°C\)

Step 1: Heat gained by water
\[q_{\text{water}} = m c \Delta T = (170)(4.18)(22 - 20)\]
\[q_{\text{water}} = (170)(4.18)(2) = 1421.2 \text{ J}\]

Step 2: Heat lost by metal = Heat gained by water
\[q_{\text{metal}} = -1421.2 \text{ J}\]

Step 3: Calculate specific heat of metal
\[q = mc\Delta T\]
\[-1421.2 = (282.25)(c)(22 - 100)\]
\[-1421.2 = (282.25)(c)(-78)\]
\[-1421.2 = -22015.5c\]
\[c = \frac{1421.2}{22015.5} = 0.0646 \text{ J/g°C}\]

Summary of All Trials:
Trial Specific Heat (J/g°C)
1 0.0646
2 0.169
3 0.0884
4 0.145
5 0.136

Problem 9

Calculate the average specific heat and density from all five trials.

Average Specific Heat = 0.120 J/g°C
Average Density = 53.2 g/cm³ (or 29.8 g/cm³ excluding outlier Trial 5)
Average Specific Heat:
\[\text{Average} = \frac{0.0646 + 0.169 + 0.0884 + 0.145 + 0.136}{5} = \frac{0.603}{5} = 0.120 \text{ J/g°C}\]

Average Density (all trials):
\[\text{Average} = \frac{14.1 + 30.5 + 35.8 + 38.9 + 146.9}{5} = \frac{266.2}{5} = 53.2 \text{ g/cm}^3\]

Average Density (excluding Trial 5 outlier):
\[\text{Average} = \frac{14.1 + 30.5 + 35.8 + 38.9}{4} = \frac{119.3}{4} = 29.8 \text{ g/cm}^3\]

Problem 10

Suggest, with a reason/reference to your lab work, a possible identity for the unknown metal.

The unknown metal is most likely Lead.
Reasoning:
Comparing experimental values to known metals:

Metal Density (g/cm³) Specific Heat (J/g°C)
Experimental ~29.8 (excl. outlier) 0.120
Lead 11.3 0.128
Copper 8.96 0.385
Iron 7.87 0.449
Aluminum 2.70 0.89

The specific heat of 0.120 J/g°C is closest to lead's 0.128 J/g°C. The density values show some experimental error, possibly due to volume measurement difficulties with small samples.

Problem 11

Calculate the percent error for the experimental specific heat and density determination (assuming lead).

Percent Error for Specific Heat = 6.3%
Percent Error for Density = 164% (with outlier) or 164% (without outlier, suggesting systematic error)
Specific Heat Percent Error:
\[\text{Percent Error} = \frac{|0.120 - 0.128|}{0.128} \times 100\% = \frac{0.008}{0.128} \times 100\% = 6.3\%\]

Density Percent Error (excluding outlier):
\[\text{Percent Error} = \frac{|29.8 - 11.3|}{11.3} \times 100\% = \frac{18.5}{11.3} \times 100\% = 164\%\]

Analysis: The specific heat determination is quite accurate (6.3% error), but the density shows large error. This suggests difficulties in accurately measuring small volumes of the metal samples.

Section C Additional Practice Problems - Answers

Problem 12

A 45.0 g sample of water is heated from 15.0°C to 75.0°C. How much heat energy was absorbed?

Heat absorbed = 11,300 J or 11.3 kJ
Given: m = 45.0 g, c = 4.18 J/g°C, \(T_i = 15.0°C\), \(T_f = 75.0°C\)
Find: q

\[\Delta T = T_f - T_i = 75.0 - 15.0 = 60.0°C\]
\[q = mc\Delta T = (45.0)(4.18)(60.0) = 11,286 \text{ J} = 11,300 \text{ J}\] (3 sig figs)

Problem 13

A 125 g piece of iron at 200°C is placed in 500 g of water at 25°C. What is the final temperature?

Final temperature = 29.7°C
Given:
Iron: m = 125 g, c = 0.449 J/g°C, \(T_i = 200°C\)
Water: m = 500 g, c = 4.18 J/g°C, \(T_i = 25°C\)
Both reach \(T_f\)

Heat lost by iron = Heat gained by water:
\[-(125)(0.449)(T_f - 200) = (500)(4.18)(T_f - 25)\]
\[-(56.125)(T_f - 200) = (2090)(T_f - 25)\]
\[-56.125T_f + 11,225 = 2090T_f - 52,250\]
\[11,225 + 52,250 = 2090T_f + 56.125T_f\]
\[63,475 = 2146.125T_f\]
\[T_f = 29.6°C \approx 29.7°C\]

Problem 14

An unknown metal (85.0 g at 100°C) is placed in 150 g of water at 22.0°C. Final temperature is 28.5°C. Calculate the specific heat.

Specific heat of metal = 0.67 J/g°C
Heat gained by water:
\[q_{\text{water}} = (150)(4.18)(28.5 - 22.0) = (150)(4.18)(6.5) = 4075.5 \text{ J}\]

Heat lost by metal:
\[q_{\text{metal}} = -4075.5 \text{ J}\]
\[-4075.5 = (85.0)(c)(28.5 - 100)\]
\[-4075.5 = (85.0)(c)(-71.5)\]
\[-4075.5 = -6077.5c\]
\[c = \frac{4075.5}{6077.5} = 0.671 \text{ J/g°C} = 0.67 \text{ J/g°C}\]

Problem 15

How much heat is released when 250 g of aluminum cools from 150°C to 25°C?

Heat released = 27,800 J or 27.8 kJ
Given: m = 250 g, c = 0.89 J/g°C, \(T_i = 150°C\), \(T_f = 25°C\)

\[\Delta T = T_f - T_i = 25 - 150 = -125°C\]
\[q = mc\Delta T = (250)(0.89)(-125) = -27,812.5 \text{ J}\]

The negative sign indicates heat is released (exothermic).
Heat released = 27,800 J = 27.8 kJ (3 sig figs)

Continue Learning

Back to Quiz Back to Lesson