Question

If 500 kcal of heat is removed from 5 tons of potatoes having specific heat of 0.1 kcal/kg \(^{\circ}\text{C}\); the temperature of the produce reduces by -

• A. 5 $^{\circ}$C
• B. 0.1 $^{\circ}$C
• C. 1 $^{\circ}$C
• D. 10 $^{\circ}$C

Hint:

Always convert mass from tons to kilograms first (\(1\text{ ton} = 1000\text{ kg}\)) to ensure dimensional compatibility with the units of specific heat capacity (kcal/kg\(\cdot^{\circ}\text{C}\)).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The temperature change of an agricultural commodity during cooling or heating depends on the amount of heat energy transferred, the mass of the commodity, and its specific heat capacity.
Specific heat capacity is the amount of heat required to change the temperature of a unit mass of substance by one degree.
Key Formula or Approach:
The sensible heat transfer equation is expressed as:
\[ Q = m \cdot c_p \cdot \Delta T \]
where:
\(Q\) is the heat removed (kcal),
\(m\) is the mass of the product (kg),
\(c_p\) is the specific heat capacity (kcal/kg\(\cdot^{\circ}\text{C}\)),
\(\Delta T\) is the temperature reduction (\(^{\circ}\text{C}\)).

Step 2: Detailed Explanation:
Let us identify and convert all given variables into standard units:
Given heat removed, \(Q = 500\text{ kcal}\).
Mass of potatoes, \(m = 5\text{ tons}\).
Since \(1\text{ ton} = 1000\text{ kg}\), the mass in kilograms is:
\[ m = 5 \times 1000 = 5000\text{ kg} \]
Specific heat capacity, \(c_p = 0.1\text{ kcal/kg}\cdot^{\circ}\text{C}\).
Substituting these values into the heat transfer equation:
\[ 500 = 5000 \times 0.1 \times \Delta T \]
Simplifying the right side of the equation:
\[ 500 = 500 \times \Delta T \]
Solving for \(\Delta T\):
\[ \Delta T = \frac{500}{500} = 1^{\circ}\text{C} \]
Thus, the temperature of the potatoes decreases by exactly \(1^{\circ}\text{C}\).

Step 3: Final Answer:
The temperature of the produce reduces by 1 $^{\circ}$C.