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How much heat (in cal) is necessary to heat 20.0 grams of octane (s = 0.526 cal/g °C) from 50.6°C to 67.2°C?

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The heat necessary to raise the temperature of 20.0 grams of octane from 50.6°C to 67.2°C is 174.5 calories.

The heat required to heat 20.0 grams of octane from 50.6°C to 67.2°C can be calculated using the specific heat formula q = mcΔT, where q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Here, m = 20.0 g (mass of octane), c = 0.526 cal/g°C (specific heat capacity of octane), and ΔT = 67.2°C - 50.6°C.

First, calculate the temperature change:ΔT = 67.2°C - 50.6°C = 16.6°C. Now, calculate the heat energy:
q = 20.0 g × 0.526 cal/g°C × 16.6°C.

After performing the multiplication, heat (q) required is 174.5 cal.

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