Difficulty: Easy
Correct Answer: 1
Explanation:
Introduction / Context:
Thermal property data appear in different unit systems. Converting specific heat capacities between FPS and SI/metric units is a routine task in energy balances.
Given Data / Assumptions:
Concept / Approach:
Start from 1 BTU/lb·°F and convert numerator (energy), then denominator (mass and temperature interval) to kcal/kg·°C.
Step-by-Step Solution:
Express in SI: 1 BTU/lb·°F = 1055.06 J / (0.453592 kg * 5/9 °C).Compute denominator factor: 0.453592 * 5/9 ≈ 0.252.Value ≈ 1055.06 / 0.252 ≈ 4187 J/kg·°C.Convert to kcal/kg·°C: 4187 / 4184 ≈ 1.00.
Verification / Alternative check:
Common reference: 1 BTU/lb·°F ≈ 4.186 kJ/kg·°C ≈ 1 kcal/kg·°C.
Why Other Options Are Wrong:
Values like 2.42, 1.987, or 4.97 are inconsistent with the precise conversion and would severely misstate heat loads.
Common Pitfalls:
Forgetting to convert °F to °C intervals; mixing mass conversions.
Final Answer:
1
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