Math Problem Statement

Calculate the energy required to heat 2 kg of water from 20°C to 80°C, given the specific heat of water is 4.18 kJ/kg·K. 418 kJ 502 kJ 668 kJ 836 kJ

Solution

The formula to calculate the energy required to heat a substance is:

Q=mcΔTQ = mc\Delta T

Where:

  • QQ is the heat energy (in kJ),
  • mm is the mass of the substance (in kg),
  • cc is the specific heat capacity (in kJ/kg·K),
  • ΔT\Delta T is the change in temperature (in °C or K).

Given:

  • m=2kgm = 2 \, \text{kg},
  • c=4.18kJ/kg\cdotpKc = 4.18 \, \text{kJ/kg·K},
  • ΔT=80°C20°C=60°C\Delta T = 80°C - 20°C = 60°C.

Now, substituting the values:

Q=2kg×4.18kJ/kg\cdotpK×60KQ = 2 \, \text{kg} \times 4.18 \, \text{kJ/kg·K} \times 60 \, \text{K}

Q=2×4.18×60=502.8kJQ = 2 \times 4.18 \times 60 = 502.8 \, \text{kJ}

So, the energy required is approximately 502 kJ.

The correct answer is 502 kJ.


Would you like more details on the calculation or have any further questions?

Here are some related questions you might find interesting:

  1. How does the specific heat capacity of different materials affect the energy required to heat them?
  2. What would happen to the energy required if the mass of the water was doubled?
  3. How does the energy required to heat water compare to other liquids, like alcohol or oil?
  4. How is the concept of specific heat capacity used in real-world applications like climate control systems?
  5. What would be the energy required if the temperature change was from 10°C to 100°C?

Tip: The specific heat capacity is a property that varies from substance to substance and can even change under different conditions (e.g., pressure or phase changes).

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Heat Transfer
Unit Conversion

Formulas

Q = mcΔT

Theorems

First Law of Thermodynamics

Suitable Grade Level

Grades 9-12