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?

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

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.

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