Problem Statement
How much heat is required to raise the temperature of $0.50\,\text{kg}$ of water by $30\,\text{K}$? Take the specific heat of water as $4186\,\text{J/(kg·K)}$. (HC Verma, Chapter 25 — Calorimetry, Problem 38.)
Given Information
- $m = 0.50\,\text{kg}$ — mass of water
- $c = 4186\,\text{J/(kg·K)}$ — specific heat of water
- $\Delta T = 30\,\text{K}$ — required temperature rise
Physical Concepts & Formulas
Heat exchanged by a body changes its temperature in proportion to its mass and specific heat, or drives a phase change at constant temperature through the latent heat. In an isolated system heat lost equals heat gained.
- $Q=mc\,\Delta T$ — sensible heat to change temperature
- $Q=mL$ — latent heat for a phase change
- $Q_\text{lost}=Q_\text{gained}$ — calorimetric energy balance
Step-by-Step Solution
Step 1 — Choose the calorimetry relation: No phase change occurs, so the heat needed is purely sensible heat.
$$Q=mc\,\Delta T$$
Step 2 — Substitute the data: Insert the mass, specific heat of water and the temperature rise.
$$Q=(0.50)(4186)(30)$$
Step 3 — Evaluate the heat required: Multiply the three factors to find the energy input.
$$Q=6.28\times10^{4}\,\text{J}$$
Worked Calculation
$$Q=(0.50)(4186)(30)=6.28\times10^{4}\,\text{J}$$
Answer
$$\boxed{Q\approx6.3\times10^{4}\,\text{J}}$$
About 63 kilojoules are needed to warm half a kilogram of water by 30 K.
Physical Interpretation
Water’s large specific heat is why it stores so much energy; this magnitude is consistent with the few minutes a kettle takes to heat a similar amount of water.
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