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A 850 kg cube shaped box made of steel is placed in a deep part of the ocean with 450 celcius water and the box starts at 15 celcius. If the rate of energy transfer is 29.45 x 10^3 W, then how many hours does it need to be in the water to get to 450 celcius?

User Nayef Harb
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1 Answer

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Answer:

The time it will take to raise the temperature of 850 kg of steel from 15°C to 45°C is approximately 0.11064 hours

Step-by-step explanation:

The given parameters are;

The mass of the steel = 850 kg

The temperature of the water of the ocean = 45°C

The temperature of the steel = 15°C

The heat capacity for steel, c = 0.46 kJ/(kg·K)

Therefore, the heat, ΔQ required to raise the temperature of 850 kg of steel from 15°C to 45°C is given as follows;

ΔQ = The mass of steel × The heat capacity for steel × The rise in temperature

ΔQ = 850 kg ×0.46 kJ/(kg·K) × (45 - 15)°C = 11730 kJ = 11730 × 10³J

The rate of heat transfer = 29.45 × 10³ W = 29.45 × 10³ J/S

Therefore, the time, t, it will take to absorb 11730 kJ is given as follows;

t = (11730 × 10³J)/(29.45 × 10³ J/S) ≈ 398.3 seconds

t ≈ 398.3 seconds = 398.3 seconds ×1/60 minutes/second × 1/60 hour/minute

t ≈ 0.11064 hours

The time it will take to raise the temperature of 850 kg of steel from 15°C to 45°C ≈ 0.11064 hours.

User Wayne Wei
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