Question

gap region between the diver's skin and her
wetsuit, forming a water layer about 0.5 mm thick.
Assume that the total surface area of the wetsuit
covering the diver is about 1.0 mº, and that the
water enters the suit at 10 °C and is warmed by
the diver to skin temperature of 35°C. The specific
heat of water is 1.00 kcal/kg - Cº.
Estimate how much energy (in
units of candy bars = 300 kcal)
is required by this heating
process.
Express your answer using
two significant figures.

Answer 1

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

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