By the end of this chapter, you will be able to:
Mastering these skills will help you ensure precise temperature control, which is essential for quality and safety in many technical trades.
Thermometric quantities such as temperature and heat are fundamental in Science Laboratory Technology, especially in Kenya where precise measurement and control of thermal conditions affect experimental accuracy and industrial processes. This chapter focuses on understanding heat and temperature, then explores various instruments used to measure temperature reliably. Mastery of these concepts enables laboratory technologists to maintain controlled environments, calibrate equipment correctly, and ensure quality in scientific investigations.
Heat and temperature are closely related physical quantities but represent different concepts. Heat refers to energy transferred due to temperature differences, while temperature measures the thermal state of a body or system. Both are critical in laboratory settings for controlling reactions, preserving samples, and conducting experiments under defined thermal conditions.
Heat energy is the form of energy transferred between systems or bodies due to a temperature difference. It is measured in joules (J) or calories (cal). The relationship between heat energy \(Q\), mass \(m\), specific heat capacity \(c\), and temperature change \(\Delta T\) is fundamental:
$$Q = mc\Delta T$$
Example 1: A laboratory beaker containing 500 g of water is heated, causing its temperature to rise from 20°C to 80°C. Calculate the heat energy absorbed by the water. Given: \(c = 4.18 \, \text{kJ/kg}^\circ C\).
Given:
\(m = 0.5 \, \text{kg}\), \(c = 4180 \, \text{J/kg}^\circ C\), \(\Delta T = 80 - 20 = 60^\circ C\)
$$Q = mc\Delta T$$
$$Q = 0.5 \times 4180 \times 60$$
$$Q = 125,400 \, \text{J}$$
Answer: 125,400 J
Example 2: A copper block of mass 2 kg is cooled from 100°C to 25°C. If the specific heat capacity of copper is 390 J/kg°C, find the heat lost by the block.
Given:
\(m = 2 \, \text{kg}\), \(c = 390 \, \text{J/kg}^\circ C\), \(\Delta T = 100 - 25 = 75^\circ C\)
$$Q = mc\Delta T$$
$$Q = 2 \times 390 \times 75$$
$$Q = 58,500 \, \text{J}$$
Answer: 58,500 J
Example 3: A chemical reaction in a calorimeter raises the temperature of 250 g of water from 22°C to 28°C. Calculate the heat released by the reaction. Specific heat capacity of water is 4.18 kJ/kg°C.
Given:
\(m = 0.25 \, \text{kg}\), \(c = 4180 \, \text{J/kg}^\circ C\), \(\Delta T = 28 - 22 = 6^\circ C\)
$$Q = mc\Delta T$$
$$Q = 0.25 \times 4180 \times 6$$
$$Q = 6,270 \, \text{J}$$
Answer: 6,270 J
Temperature quantifies the degree of hotness or coldness of a substance. It is a scalar quantity measured in degrees Celsius (°C), Kelvin (K), or Fahrenheit (°F). The Kelvin scale is absolute and widely used in scientific calculations.
Temperature scales conversion formulas:
Example 1: Convert 25°C to Kelvin.
Given: \(T_C = 25^\circ C\)
$$T_K = T_C + 273.15$$
$$T_K = 25 + 273.15$$
$$T_K = 298.15 \, K$$
Answer: 298.15 K
Example 2: Convert 100°C to Fahrenheit.
Given: \(T_C = 100^\circ C\)
$$T_F = \frac{9}{5} T_C + 32$$
$$T_F = \frac{9}{5} \times 100 + 32$$
$$T_F = 180 + 32$$
$$T_F = 212^\circ F$$
Answer: 212°F
Example 3: A sample is at 310 K. Convert this temperature to Celsius.
Given: \(T_K = 310 \, K\)
$$T_C = T_K - 273.15$$
$$T_C = 310 - 273.15$$
$$T_C = 36.85^\circ C$$
Answer: 36.85°C
Heat transfer causes temperature changes in substances, but the relationship depends on the material's properties, especially specific heat capacity. For the same amount of heat, materials with low specific heat change temperature more than those with high specific heat.
The formula \(Q = mc\Delta T\) relates heat energy to temperature change, where \(Q\) is heat energy, \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is temperature difference.
Example 1: Which substance requires more heat to raise the temperature of 1 kg by 10°C: water (\(c=4180\)) or aluminum (\(c=900\))?
Calculate \(Q\) for both:
Water:
$$Q = 1 \times 4180 \times 10 = 41,800 \, \text{J}$$
Aluminum:
$$Q = 1 \times 900 \times 10 = 9,000 \, \text{J}$$
Answer: Water requires 41,800 J, aluminum 9,000 J; water requires more heat.
Example 2: A metal block of mass 3 kg absorbs 27,000 J of heat and its temperature rises by 15°C. Find the specific heat capacity of the metal.
Given:
\(Q=27000 \, J\), \(m=3 \, kg\), \(\Delta T=15^\circ C\)
Formula:
$$c = \frac{Q}{m\Delta T}$$
Substitution:
$$c = \frac{27000}{3 \times 15}$$
$$c = \frac{27000}{45}$$
$$c = 600 \, \text{J/kg}^\circ C$$
Answer: 600 J/kg°C
Example 3: Calculate the temperature rise of 2 kg of water when 10,000 J of heat is added.
Given:
\(Q=10,000 \, J\), \(m=2 \, kg\), \(c=4180 \, \text{J/kg}^\circ C\)
Formula:
$$\Delta T = \frac{Q}{mc}$$
Substitution:
$$\Delta T = \frac{10,000}{2 \times 4180}$$
$$\Delta T = \frac{10,000}{8360}$$
$$\Delta T = 1.196^\circ C$$
Answer: 1.196°C
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Create a free accountThis chapter explored the fundamental concepts of heat and temperature, distinguishing between the two and explaining their significance in physical measurements. Various temperature measuring instruments were examined, highlighting their principles and applications in different contexts. The chapter then detailed the processes involved in changes of state, including melting, vaporisation, and sublimation or deposition, providing a clear understanding of phase transitions. Methods for determining specific heat and latent heats of fusion and vaporisation were discussed, emphasizing their importance in thermodynamic calculations. The concept of thermometric quantities was introduced, explaining how measurable physical properties relate to temperature changes. Finally, guidance on report writing for practical experiments involving the measurement and control of thermometric quantities was provided, ensuring accurate documentation and analysis. Together, these topics form a comprehensive foundation for understanding and managing thermal measurements in engineering practice.
A metal block of mass 2 kg is heated from 20°C to 80°C. If the specific heat capacity of the metal is \( 400 \, \text{J/kg}^\circ\text{C} \), calculate the heat energy absorbed by the block. (2 marks)
A laboratory thermometer shows a temperature of 350 K. Convert this temperature to degrees Celsius. (2 marks)
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