What is the heat in joules required to raise the temperature of 25 grams of water from 0 degrees C to 100 degrees C in calories?

Text Solution

Solution : Given m=25 g , `DeltaT`=(100-0)=`100^@C` <br> or in terms of Kelvin (373.15-273.15)=100K, `C=4.18J//g^@C` <br> Heat energy required , Q=m x C x `DeltaT` =25 x 4.18 x 100 = 10450 J

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")`

Given m = 25 g, ∆T = (100 – 0) = 100°COr 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

Concept: Specific Heat Capacity

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