Question

A girl in space has managed to lay claim to an abandoned freighter whose crew was killed by Cubarb parasites. She fired the engines of her craft for 30 seconds, which involved igniting unobtainium gas to propel the ship (in an action-reaction process). During this time, the engine cones absorbed some of the heat of this explosion. If one can approximate the engine cones as four slabs of 700 kg steel, 70000 kJ of heat was absorbed, and the initial temperature of the steel was -250 Celsius, what is the final temperature? Is the steel in danger of melting?

Answer 1

We know that, heat gained = m*S*deltaT where m is mass, S is specific heat, deltaT is temperature change.

70000e3 = 4*700*490*(T- -250)

T = -199 degree C answer

No, it is not in the danger of melting because final temperature is still too less (even less than melting point of water )

Answer 2

  Calculation:

1. Key Data:

• Mass of steel (m): 4 slabs × 700 kg = 2,800 kg

• Heat absorbed (Q): 70,000 kJ = 70,000,000 J

• Specific heat capacity of steel (c): 450 J/kg·°C

• Initial temperature (T₀): -250°C

• Melting point of steel: ~1,370–1,530°C

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2. Calculate Temperature Change (ΔT):

$$Q=mc\mathrm{\Delta }T⟹\mathrm{\Delta }T=\frac{Q}{mc}=\frac{70,000,000}{2,800\times 450}=55.56\mathrm{°}C$$

3. Final Temperature (T):

$$T=T_0+\mathrm{\Delta }T=-250\mathrm{°}C+55.56\mathrm{°}C=-194.44\mathrm{°}C$$

Wait, this doesn’t make sense!

• The heat absorbed (70,000 kJ) is huge, but the calculation suggests only a +55.56°C rise.

• 
  

• Mistake: The formula assumes all heat goes into temperature change, but steel’s phase changes (melting) absorb energy without raising temperature.

• 
  

4. Revised Approach (Including Latent Heat):

1. Heat to warm steel from -250°C to 0°C:

   $$Q_1=mc\mathrm{\Delta }T=2,800\times 450\times (0-(-250))=315,000,000\text{J}$$

• But only 70,000,000 J is available → Not enough to reach 0°C!

2. Actual Temperature Rise:

   $$\mathrm{\Delta }T=\frac{70,000,000}{2,800\times 450}=55.56\mathrm{°}C$$$$T=-250\mathrm{°}C+55.56\mathrm{°}C=-194.44\mathrm{°}C$$

Conclusion:

• The final temperature (-194.44°C) is still far below melting.

• Error in Problem Statement: 70,000 kJ is too low to pose a melting risk.