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Question
Numerical Problem.
What is the heat in joules required to raise the temperature of 25 grams of water from 0°C to 100°C? What is the heat in Calories? (Specific heat of water = `(4.18"J")/("g"°"C")`
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Solution
Given m = 25 g, ∆T = (100 – 0) = 100°C
Or in terms of Kelvin (373.15 – 273.15) = 100K,
C = `(4.18"J")/("g"°"C")`
Heat energy required, Q = m × C × ∆T = 25 × 4.18 × 100 = 10450 J
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