31. An amount of ice of mass 10^-3 kg and temperature -10°C is transformed to vapour of temperature 110°C by applying heat. The total amount of work required for this conversion is: Take: Specific heat of ice = 2100 J/kg·K, Specific heat of water = 4180 J/kg·K, Specific heat of steam = 1920 J/kg·K, Latent heat of ice = 3.35×10^5 J/kg, Latent heat of steam = 2.25×10^6 J/kg. Option 1: 3022 J Option 2: 3043 J Option 3: 3003 J Option 4: 3024 J

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What is the question?

We have a teeny-tiny piece of ice (just 1 gram!) that starts out really cold at –10 °C. We keep warming it up until it finally becomes hot steam at 110 °C.

Why is it tricky?

Ice does four different things while warming:

1. Gets less cold (–10 °C → 0 °C)
2. Melts (turns to water)
3. Gets hotter (0 °C → 100 °C as liquid)
4. Boils (turns to steam)
5. Gets even hotter (100 °C → 110 °C as steam)

Each of these steps needs its own amount of heat. Add all those heats together and you know how much energy ("work" in the question) you have to supply.

That’s all the question is asking: “Add up the heat for each step.”

Step-by-Step Calculation

Given:
$m = 10^{-3}$ kg

Constants:
$c_{\text{ice}} = 2100$ J kg⁻¹ K⁻¹
$c_{\text{water}} = 4180$ J kg⁻¹ K⁻¹
$c_{\text{steam}} = 1920$ J kg⁻¹ K⁻¹
$L_f = 3.35 \times 10^5$ J kg⁻¹
$L_v = 2.25 \times 10^6$ J kg⁻¹

1. Warm ice from –10 °C to 0 °C

2. Melt ice at 0 °C

3. Warm water from 0 °C to 100 °C

4. Vaporise water at 100 °C

5. Warm steam from 100 °C to 110 °C

6. Total Heat (Work) Required

Rounded to the nearest joule, $Q_{\text{total}} \approx 3043\,\text{J}$

Hence, the correct option is Option 2: 3043 J.