Chapter 10: Heat Capacity (Numerical Problems)

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Chapter Overview & SLOs

How is the thermal energy lost or gained by a substance calculated? Using the fundamental formula Q = mcΔT, where m is mass (kg or g), c is specific heat capacity (J/kg°C or J/g°C), and ΔT is the change in temperature (°C).

What is the specific heat capacity of water? Water has a specific heat capacity of 4186 J/kg°C or 4.18 J/g°C. This high value means water requires a large amount of heat to change temperature, making it an ideal coolant.

How do we determine the specific heat capacity of an unknown material? By rearranging the heat formula to c = Q/mΔT, we can calculate the capacity if the energy absorbed, mass, and temperature shift are known.

What is the Principle of Conservation of Energy in calorimetry? The principle states that heat lost by a hotter substance equals the heat gained by the cooler substance and its container. Heat lost by hot object = Heat gained by cold water + Heat gained by calorimeter.

How do we find the final temperature (Tf) of a water mixture? By setting m₁c(T₁ - Tf) = m₂c(Tf - T₂), we can solve for Tf. Note: The specific heat capacity c cancels out when both substances are the same (water).

Example - Water mixture final temperature: Mixing 100 g water at 80°C with 200 g water at 20°C: 100 × (80 - Tf) = 200 × (Tf - 20) → 8000 - 100Tf = 200Tf - 4000 → 12000 = 300Tf → Tf = 40°C.

How do we calculate heat required to raise temperature of a large mass of water? For 25 kg of water with a 50°C rise: Q = mcΔT = 25 × 4186 × 50 = 5,232,500 J = 5.23 MJ.

Unit conversions: 1 kJ = 1000 J, 1 MJ = 1,000,000 J. To convert J to kJ, divide by 1000; to convert J to MJ, divide by 1,000,000.

Volume to mass conversion for water: Since the density of water is 1 kg/L, 1 liter of water has a mass of 1 kg. Volume in liters × 1 kg/L = mass in kg.

Sign convention: Heat lost results in a negative value for Q, while heat gained is positive.

These notes are strictly aligned with the Student Learning Outcomes (SLOs) for the FBISE 2026 annual examination.

• How do we convert specific heat units from J kg⁻¹ K⁻¹ to J g⁻¹ °C⁻¹? Since 1 kg = 1000 g and a change of 1 Kelvin equals 1 degree Celsius, the value is divided by 1000; for example, water's specific heat becomes 4.18 J g⁻¹ °C⁻¹.
• How do we calculate the heat required to raise the temperature of a large mass of water? For 25 kg of water (c = 4186 J kg⁻¹ °C⁻¹) experiencing a 50°C rise, the energy required is approximately 5.23 MJ using $Q = mc\Delta T$.
• How do we find the final temperature (Tf) of a water mixture? By setting $m_1c(T_1 - T_f) = m_2c(T_f - T_2)$, we can solve for $T_f$; for instance, mixing 100 g water at 80°C with 200 g water at 20°C results in a final temperature of 40°C.
• How is calorimetry used to find the specific heat ($c_s$) of a solid? The heat lost by the solid ($m_s c_s \Delta T_s$) is equated to the sum of heat gained by the water ($m_w c_w \Delta T_w$) and the calorimeter ($m_c c_c \Delta T_c$), allowing us to isolate and solve for $c_s$.

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