Answer:
∴ 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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