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Answer :
The quantity of energy transferred is equal to 18,485,376 Joules.
Given the following data:
- Mass of water = 16 kg
- Specific heat capacity of water = c = 4186 J/kgK
- Temperature = 3°C
Conversion:
Temperature = 3°C to Kelvin = [tex]273+3=276\;K[/tex]
To calculate the quantity of energy transferred:
Mathematically, quantity of energy is given by the formula;
[tex]Q = mct[/tex]
Where:
- Q represents the quantity of energy.
- m represents the mass of an object.
- c represents the specific heat capacity.
- t represents the temperature.
Substituting the given parameters into the formula, we have;
[tex]Q = 16\times 4186 \times 276[/tex]
Quantity of energy, Q = 18,485,376 Joules
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Final answer:
To calculate the energy transferred when heating 16 kg of water by 3°C, use the formula Q = mcΔT, leading to an answer of 2.0 x 10^5 J.
Explanation:
The question asks to calculate the energy transferred when raising the temperature of 16 kg of water by 3°C, with a given specific heat capacity of water as 4,186 J/kg·K. The formula to calculate the heat energy (Q) transferred in changing the temperature of a substance is Q = mcΔT, where m is the mass of the substance (in kg), c is the specific heat capacity (in J/kg·K), and ΔT is the change in temperature (in °C or K).
Substituting the given values: Q = 16 kg * 4,186 J/kg·K * 3 °C = 200,616 J, which simplifies to 2.01 × 105 J or 201.616 kJ.
Therefore, the correct answer is d. 2.0 × 105 J.
