1.1 Specific Heat Capacity

Specific heat capacity (c) is the quantity of heat required to raise the temperature of 1 kg of a substance by 1°C. Its unit is J/kg°C (joules per kilogram degree Celsius).

The heat equation relates heat transferred to mass, specific heat capacity, and temperature change:

$H = m \times c \times \Delta T$

Where:

• H = heat gained or lost (J)
• m = mass (kg)
• c = specific heat capacity (J/kg°C)
• ΔT = change in temperature (°C)

1.2 Worked Example: Heating Water in a Tanzanian Kitchen

A household in Dar es Salaam heats 2 kg of water from 25°C to 85°C using an electric kettle. The specific heat capacity of water is 4,200 J/kg°C.

Given:

• Mass, m = 2 kg
• Initial temperature, Tᵢ = 25°C
• Final temperature, T_f = 85°C
• Specific heat capacity, c = 4,200 J/kg°C

Required: Heat supplied (H)

Solution:

$\Delta T = T_f - T_i = 85°C - 25°C = 60°C$

$H = m \times c \times \Delta T$ $H = 2 \times 4200 \times 60$ $H = 504,000 \, \text{J} = 504 \, \text{kJ}$

The kettle supplies 504 kJ of heat energy to the water.

1.3 Principle of Heat Exchange (Calorimetry)

When hot and cold objects are placed together, heat flows from the hot object to the cold one until they reach thermal equilibrium. By the law of conservation of energy:

$\text{Heat lost by hot object} = \text{Heat gained by cold object}$

This principle allows us to solve problems involving mixtures of substances at different temperatures.

2.1 Ohm's Law

Ohm's Law states that the potential difference (voltage) across a conductor is directly proportional to the current flowing through it, provided temperature remains constant:

$V = I \times R$

Where:

• V = voltage (V, volts)
• I = current (A, amperes)
• R = resistance (Ω, ohms)

2.2 Worked Example: Calculating Current in a Simple Circuit

A student connects a small radio to a 12V car battery in Arusha. The radio has a resistance of 24 Ω. What current flows through the radio?

Given:

• Voltage, V = 12 V
• Resistance, R = 24 Ω

Required: Current (I)

Solution:

$I = \frac{V}{R} = \frac{12}{24} = 0.5 \, \text{A}$

A current of 0.5 amperes flows through the radio.

2.3 Resistance of a Wire

The resistance of a uniform wire depends on its length and cross-sectional area:

$R = \rho \frac{L}{A}$

Where:

• ρ (rho) = resistivity of the material (Ω·m)
• L = length of wire (m)
• A = cross-sectional area (m²)

Key relationships:

• Doubling the length doubles the resistance
• Doubling the cross-sectional area halves the resistance

2.4 Combining Resistors

In Series: $R_{eq} = R_1 + R_2 + R_3$

In Parallel: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

3.1 Solar Energy Overview

Solar energy is energy from the Sun, captured using solar cells (photovoltaic cells) that convert sunlight directly into electricity. Solar panels are increasingly used in Tanzania for homes, schools, and businesses.

3.2 Power Output of a Solar Panel

The electrical power generated by a solar panel can be calculated using:

$P = V \times I$

Or, using Ohm's Law:

$P = I^2 \times R = \frac{V^2}{R}$

Where:

• P = power (W, watts)
• V = voltage (V)
• I = current (A)

3.3 Worked Example: Solar Panel in Mwanza

A solar panel in Mwanza produces 5 A of current at 18 V when sunlight is strong.

Given:

• Current, I = 5 A
• Voltage, V = 18 V

Required: Power output (P)

Solution:

$P = V \times I = 18 \times 5 = 90 \, \text{W}$

The solar panel delivers 90 watts of electrical power, which can run small appliances or charge batteries for later use.

3.4 Energy Stored in Batteries

Energy stored in a battery is calculated as:

$E = P \times t$

Or:

$E = V \times I \times t$

Where:

• E = energy (Wh, watt-hours)
• t = time (hours)

In Tanzania, many rural villages without grid electricity use solar home systems to power lights and charge mobile phones. For example, a family in a village near Mbeya might install a 50-watt solar panel that operates for 5 hours daily, generating 250 Wh of energy. Using the formula E = P × t, they can calculate how long their battery will last and plan their energy use efficiently. This application of physics helps improve daily life, supports education (students can study after dark), and reduces dependence on expensive diesel generators or charcoal.