Determine the amount of heat that would have to be applied to a 50g block of pure ice at -150*c to steam at 150.0*c

Q = 30750 j

Explanation:

Given data:

Mass of block = 50 g

Initial temperature = -150.0 °C

Final temperature = 150.0 °C

Amount of heat required = ?

Solution:

Specific heat capacity of ice = 2.05 j/g. °C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T₂- T₁

ΔT = 150 - (-150)

ΔT = 300 °C

Q = m.c. ΔT

Q = 50 g × 2.05 j/g. °C × 300 °C

Q = 30750 j

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