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Sagot :
The energy required by the heating process is 0.042 candy bars.
We'll begin by calculating the mass of the water.
- Thickness = 0.5 mm = 0.5 / 1000 = 5×10¯⁴ m
- Area = 1 m²
- Volume = Area × Thickness = 1 × 5×10¯⁴ = 5×10¯⁴ m³
- Density of water = 1000 Kg/m³
- Mass of water =?
Mass = Density × Volume
Mass of water = 1000 × 5×10¯⁴
Mass of water = 0.5 Kg
Next, we shall determine the heat required
- Mass of water (M) = 0.5 Kg
- Initial temperature of water (T₁) = 10 °C
- Final temperature (T₂) = 35 °C
- Change in temperature (ΔT) = T₂ – T₁ = 35 – 10 = 25 °C
- Specific heat capacity of water (C) = 1 Kcal/KgºC
- Heat (Q) =?
Q = MCΔT
Q = 0.5 × 1 × 25
Q = 12.5 Kcal
Finally, we shall convert 12.5 Kcal to candy bar.
300 Kcal = 1 candy bar
Therefore,
12.5 Kcal = 12.5/300
12.5 Kcal = 0.042 candy bars
Therefore, the energy required in candy bar is 0.042 candy bars
Learn more about heat transfer: https://brainly.com/question/18883151
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