Question

How much heat is required to convert 12 g of water from 15°C to 42°C ?

(Latent heat of fusion of water-80 cal/g, Latent heat of vaporization of water-540 cal/g
Specific heat of ice=0.5 cal/g. Oc and Specific heat of water- 1 cal/g.°C)
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Answer 1

The total heat required to convert 12 g of water from 15°C to 42°C is 10,164 calories.

1. **Heating the water from 15°C to 100°C:**

Calculate Q1 using the formula: Mass × Specific Heat × Temperature Change

Q1 = 12 grams × 1 cal/g.°C × 85°C = 1020 cal

2. **Melting the ice at 0°C:**

Calculate Q2 using the formula: Mass × Latent Heat of Fusion

Q2 = 12 grams × 80 cal/g = 960 cal

3. **Heating the water from 0°C to 100°C:**

Calculate Q3 using the formula: Mass × Specific Heat × Temperature Change

Q3 = 12 grams × 1 cal/g.°C × 100°C = 1200 cal

4. **Vaporizing the water at 100°C:**

Calculate Q4 using the formula: Mass × Latent Heat of Vaporization

Q4 = 12 grams × 540 cal/g = 6480 cal

5. **Heating the steam from 100°C to 42°C:**

Calculate Q5 using the formula: Mass × Specific Heat × Temperature Change

Q5 = 12 grams × 1 cal/g.°C × 42°C = 504 cal

Now, sum up all the individual heats to get the total heat required:

`\[ Q_{\text{total}} = Q1 + Q2 + Q3 + Q4 + Q5 \]`

`\[ Q_{\text{total}} = 1020 + 960 + 1200 + 6480 + 504 \, \text{cal} \]`

`\[ Q_{\text{total}} = 10,164 \, \text{cal} \]`