How many grams of butane must be burnt to heat 10 kg water from 30°C to 100°C? The heat of combustion of butane is −700 kcal/mol and the specific heat of water is 1 cal gm−1 K−1.

Explanation

Q = m x c x ΔT where, Q = heat energy required m = mass of the water c = specific heat of water ΔT = change in temperature (final temperature - initial temperature) Given values, m = 10 kg = 10,000 gm (since 1 kg = 1000 gm) c = 1 cal gm-1 K-1 ΔT = 100°C - 30°C = 70°C Plugging in the values, Q = 10,000 gm x 1 cal gm-1 K-1 x 70 K Q = 700,000 cal Now, we need to convert the heat energy from calories to grams of butane. The heat of combustion of butane is given as -700 kcal/mol. To convert this to calories per gram of butane, we need to find the molar mass of butane. The molar mass of butane (C4H10) is, Molar mass = (4 x atomic mass of carbon) + (10 x atomic mass of hydrogen) = 4 x 12.01 g/mol + 10 x 1.01 g/mol = 48.04 g/mol Now, we can calculate the heat energy released per gram of butane, -700 kcal/mol / 48.04 g/mol ≈ -14.57 kcal/g ≈ -14.57 x 1000 cal/g ≈ -14,570 cal/g Now, we can calculate the heat energy released per gram of butane, -700 kcal/mol / 48.04 g/mol ≈ -14.57 kcal/g ≈ -14.57 x 1000 cal/g ≈ -14,570 cal/g ≈ -14,570 cal/g Since the heat of combustion is given as a negative value, we use the absolute value for the calculation. So, for every 14,570 cal of heat energy released, we need 1 gram of butane. Finally, the amount of butane needed to produce 700,000 cal is, 700,000 cal / 14,570 cal/g ≈ 48.04 g ≈ 48 g The correct option is A) 58 gm.