How many kilocalories of heat are released when 75 g of steam at 100 degrees Celsius is converted to ice at 0 degrees Celsius?

∴ Q = -7.52kCal

Explanation:

Using the formula for specific heat capacity:

Q = mcΔT

where ΔT = change in temperature (final - initial)  = (0 - 100)°C = -100°C

m = mass (g) = 75g

c = specific heat capacity = 4.2 J/g°C in water

⇒ Q = 75 × 4.2 × -100

= -31,500J

But 1J - 0.000239kCal

∴ Q = -7.52kCal

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