Worksheet 2 - Specific Heat Capacity: Exercises on calculating heat energy and specific heat capacity with real-world examples.

Let's solve each exercise step by step using the formula for heat energy:

$$
Q = m \cdot c \cdot \Delta T
$$

Where:
- $ Q $ = heat energy (in joules, J)
- $ m $ = mass (in kg)
- $ c $ = specific heat capacity (in J/kg°C)
- $ \Delta T $ = change in temperature (in °C) = $ T_{\text{final}} - T_{\text{initial}} $

---

Exercise 1:

A pot is filled with 9kg of water at 10°C. Calculate how much heat energy would be needed to raise the temperature to 60°C.
[Specific heat capacity of water = 4200 J/kg°C]

Given:
- $ m = 9 \, \text{kg} $
- $ c = 4200 \, \text{J/kg°C} $
- $ \Delta T = 60^\circ\text{C} - 10^\circ\text{C} = 50^\circ\text{C} $

$$
Q = 9 \times 4200 \times 50 = 1,890,000 \, \text{J}
$$

✔ Answer: $ \boxed{1,890,000 \, \text{J}} $ or $ \boxed{1.89 \times 10^6 \, \text{J}} $

---

Exercise 2:

A girl has a bottle of mineral water (2 litres = 2 kg). The water gets heated by the sun by 5°C. How much heat has the water absorbed?

Given:
- $ m = 2 \, \text{kg} $
- $ c = 4200 \, \text{J/kg°C} $ (same as water)
- $ \Delta T = 5^\circ\text{C} $

$$
Q = 2 \times 4200 \times 5 = 42,000 \, \text{J}
$$

✔ Answer: $ \boxed{42,000 \, \text{J}} $ or $ \boxed{4.2 \times 10^4 \, \text{J}} $

---

Exercise 3:

When the temperature of 0.25 kg of ice-cream increases from -10°C to -2°C, the heat supplied is 3000 J. Find the specific heat capacity of the ice-cream.

Given:
- $ m = 0.25 \, \text{kg} $
- $ \Delta T = -2^\circ\text{C} - (-10^\circ\text{C}) = 8^\circ\text{C} $
- $ Q = 3000 \, \text{J} $

Use:
$$
c = \frac{Q}{m \cdot \Delta T} = \frac{3000}{0.25 \times 8} = \frac{3000}{2} = 1500 \, \text{J/kg°C}
$$

✔ Answer: $ \boxed{1500 \, \text{J/kg°C}} $

---

Exercise 4:

A refrigerator cools 3 liters of soft drink from 23°C to 5°C.

#### i. If density = 1015 kg/m³, find the mass.

We are given:
- Volume = 3 L = $ 3 \times 0.001 = 0.003 \, \text{m}^3 $
- Density = $ 1015 \, \text{kg/m}^3 $

$$
\text{mass} = \text{density} \times \text{volume} = 1015 \times 0.003 = 3.045 \, \text{kg}
$$

✔ Answer: $ \boxed{3.045 \, \text{kg}} $

---

#### ii. How much heat energy must be removed?

Given:
- $ m = 3.045 \, \text{kg} $
- $ c = 4150 \, \text{J/kg°C} $
- $ \Delta T = 5^\circ\text{C} - 23^\circ\text{C} = -18^\circ\text{C} $ → magnitude = 18°C (heat removed)

$$
Q = m \cdot c \cdot \Delta T = 3.045 \times 4150 \times 18
$$

First calculate:
- $ 3.045 \times 4150 = 12,632.25 $
- $ 12,632.25 \times 18 = 227,379.5 \, \text{J} $

✔ Answer: $ \boxed{227,380 \, \text{J}} $ (approx.)

---

#### iii. How long will this take if the refrigerator removes heat at 18 J/s?

Power = rate of energy removal = 18 J/s

$$
\text{Time} = \frac{\text{Energy}}{\text{Rate}} = \frac{227,380}{18} \approx 12,632.22 \, \text{seconds}
$$

Convert to minutes:
$$
12,632.22 \div 60 \approx 210.54 \, \text{minutes} \approx 3.51 \, \text{hours}
$$

✔ Answer: $ \boxed{12,632 \, \text{seconds}} $ or about $ \boxed{3.51 \, \text{hours}} $

---

✔ Final Answers Summary:

| Exercise | Answer |
|--------|--------|
| 1 | $ 1.89 \times 10^6 \, \text{J} $ |
| 2 | $ 4.2 \times 10^4 \, \text{J} $ |
| 3 | $ 1500 \, \text{J/kg°C} $ |
| 4i | $ 3.045 \, \text{kg} $ |
| 4ii | $ 227,380 \, \text{J} $ |
| 4iii | $ 12,632 \, \text{s} $ (≈ 3.51 hours) |

Let me know if you'd like these formatted for printing or need explanations in simpler terms!