Specific Heat of Corn by the Method of Mixtures: 22-Agric-A6 Physical Properties of Biological Materials and Food Products, May 2016, Q1

22-Agric-A6 Physical Properties of Biological Materials and Food Products · May 2016

Question 1: Specific Heat of Corn by the Method of Mixtures (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Method-of-mixtures calorimeter data
Quantity	Symbol	Value
Mass of thermos	$m_t$	54.5 g = 0.0545 kg
Mass of corn (12% moisture)	$m_c$	90 g = 0.090 kg
Mass of water added	$m_w$	255 g = 0.255 kg
Initial temperature, thermos + corn	$T_1$	73 °C
Initial temperature, water	$T_w$	21 °C
Equilibrium temperature	$T_e$	30 °C
Specific heat, thermos	$c_t$	0.946 kJ/(kg·K)
Specific heat, water	$c_w$	4.187 kJ/(kg·K)

Find. The specific heat of the corn, $c_c$.

Approach. Apply a heat balance across the sealed, adiabatic calorimeter: the heat given up by the thermos and the corn as they cool from $T_1$ to $T_e$ equals the heat gained by the water as it warms from $T_w$ to $T_e$.

Write the heat balance. No heat escapes the sealed thermos, so $$\left(m_t c_t + m_c c_c\right)\left(T_1 - T_e\right) = m_w c_w \left(T_e - T_w\right).$$
Compute the heat gained by the water. $$Q_w = m_w c_w \left(T_e - T_w\right) = (0.255)(4.187)(30-21) = 9.609\ \text{kJ}.$$
Compute the heat given up by the thermos itself. $$Q_t = m_t c_t \left(T_1 - T_e\right) = (0.0545)(0.946)(73-30) = 2.217\ \text{kJ}.$$ The remaining heat, $Q_c = Q_w - Q_t = 9.609 - 2.217 = 7.392\ \text{kJ}$, must have come from the corn cooling through the same 43 K.
Solve for the corn's specific heat. Substituting $m_c(T_1-T_e) = (0.090)(43) = 3.87\ \text{kg}\cdot\text{K}$, $$c_c = \frac{Q_c}{m_c\left(T_1-T_e\right)} = \frac{7.392}{3.87} = \boxed{1.91\ \text{kJ/(kg}\cdot\text{K)}}.$$

Final results
Quantity	Value
Specific heat of the corn, $c_c$	1.91 kJ/(kg·K)

Concept & Theory

Topic: Specific heat determination of a biological material by the method of mixtures (calorimetry).

Key relations: heat balance $\sum m_i c_i \Delta T_i = 0$ across an adiabatic mixing process; Siebel's empirical equation for moist food solids, $c_p \approx 1.4623 + 0.0293\,(\%\text{MC})$, in kJ/(kg·K).

Why this works: a sealed, well-insulated thermos exchanges essentially no heat with the surroundings during the short mixing interval, so the first law reduces to a simple energy balance between the hot mass (thermos + grain) and the cold mass (added water). Because the thermos and the grain are lumped together at the same initial and final temperature, their combined heat capacity can be split by first isolating the (separately known) thermos contribution, leaving the corn's contribution as the residual — exactly the "successive subtraction" logic used whenever a calorimeter itself has a non-negligible heat capacity.

Common pitfall: treating the thermos as negligible and lumping $(m_t+m_c)$ together with only the corn's specific heat, or forgetting to convert grams to kilograms consistently — both silently shift the answer. Here the thermos supplies about 23% of the total heat released, so ignoring it would understate $c_c$ by a similar margin. The Siebel check (1.81 kJ/(kg·K) at 12% moisture) is close to but not identical to the measured 1.91 kJ/(kg·K); the gap is normal for a single-run calorimeter measurement and does not indicate an error.

Source: Mohsenin, Physical Properties of Plant and Animal Materials, thermal-properties chapter (method of mixtures, Siebel's equation); Singh & Heldman, Introduction to Food Engineering, Ch. 4.

Quantity	Symbol	Value
Mass of thermos	\(m_t\)	54.5 g = 0.0545 kg
Mass of corn (12% moisture)	\(m_c\)	90 g = 0.090 kg
Mass of water added	\(m_w\)	255 g = 0.255 kg
Initial temperature, thermos + corn	\(T_1\)	73 °C
Initial temperature, water	\(T_w\)	21 °C
Equilibrium temperature	\(T_e\)	30 °C
Specific heat, thermos	\(c_t\)	0.946 kJ/(kg·K)
Specific heat, water	\(c_w\)	4.187 kJ/(kg·K)

Question 1 of 9: Specific Heat of Corn by the Method of Mixtures

Question 1: Specific Heat of Corn by the Method of Mixtures (20 marks)