Question

How much energy is required to change a 40 g ice cube from ice at −14◦c to steam at 118◦c? the specific heat of ice is 2090 j/kg · ◦ c, the specific heat of water is 4186 j/kg · ◦ c, the specific heat of stream is 2010 j/kg · ◦ c, the heat of fusion is 3.33 × 105 j/kg, and the heat of vaporization is 2.26 × 106 j/kg. answer in units of j?

Answer 1

From -44°C to 0°C: Q1 = mCΔT = 40*(2090/1000)*44 = 3678.4 J
Melting at 0°C: Q2 = mHf = 40*(3.33*10^5/1000) =13320 J
Heating liquid water from 0°C to 100°C: Q3 = mCΔT = 40*(4186/1000)*100 = 16744 J
Vaporizing water to steam at 100°C: Q4 = mHvap = 40*(2.26*10^6/100) = 90400 J
Heating steam from 100°C to 118°C: Q5 = mCΔT = 40*(2010/1000)*18 = 1447.2 J

Energy required, Q = Q1+Q2+Q3+Q4+Q5 = 3678.4+13320+16744+90400+1447.2 = 125589.6 J

Answer 2

The total amount of energy required to change 40g ice cube from -14 C to 118 C will be Q=121911 J

What is latent heat of fusion?

The amount of heat required to completely convert 0 degree ice into a

0 degree water is called as the latent heat of vaporization.

To convert ice From -14°C to 0°C:

$Q_1 = mC\Delta T = 40*(2.090 )\times 14 = 1447.2 J$

Melting at 0°C:

$Q_2 = mH_f = 40\times (3.33\times 10^5) =13320 J$

Heating liquid water from 0°C to 100°C:

$Q_3 = mC\Delta T = 40*(4.186)\times 100 = 16744 J$

Vaporizing water to steam at 100°C:

$Q_4 = mH_{vap} = 40\times \dfrac{(2.26\times 10^6)}{100} = 90400 J$

Heating steam from 100°C to 118°C:

$Q_5 = mC\DeltaT = 40\times(2.010)\times 18 = 1447.2 J$

Total energy required will be

$Q = Q1+Q2+Q3+Q4+Q5 = 1447.2+13320+16744+90400+1447.2 = 121911 J$

Thus the total amount of energy required to change 40g ice cube from -14 C to 118 C will be Q=121911 J

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