Math Problem Statement

Calculate the number of BTU’s required to change 584 lbs. of steam at 290°F to 584 lbs. of ice at 25°F.

specific heat of ice = 0.5

specific heat of water = 1.0

specific heat of steam = 0.5

latent heat of melting = 144

latent heat of boiling = 970

Solution

To calculate the number of BTUs required to convert 584 lbs of steam at 290°F into 584 lbs of ice at 25°F, we need to consider several stages of heat transfer:

  1. Cooling steam from 290°F to 212°F (condensation point).
  2. Condensing steam at 212°F into water.
  3. Cooling water from 212°F to 32°F (freezing point).
  4. Freezing water at 32°F into ice.
  5. Cooling ice from 32°F to 25°F.

We'll calculate the BTUs for each stage.

1. Cooling Steam from 290°F to 212°F

Q1=mCsteamΔT=584lbs0.5BTUlb°F(290°F212°F)Q_1 = m \cdot C_{\text{steam}} \cdot \Delta T = 584 \, \text{lbs} \cdot 0.5 \, \frac{\text{BTU}}{\text{lb} \cdot °F} \cdot (290°F - 212°F) Q1=5840.578=22752BTUQ_1 = 584 \cdot 0.5 \cdot 78 = 22752 \, \text{BTU}

2. Condensing Steam to Water at 212°F

Q2=mLatent Heat of Boiling=584lbs970BTUlbQ_2 = m \cdot \text{Latent Heat of Boiling} = 584 \, \text{lbs} \cdot 970 \, \frac{\text{BTU}}{\text{lb}} Q2=566480BTUQ_2 = 566480 \, \text{BTU}

3. Cooling Water from 212°F to 32°F

Q3=mCwaterΔT=584lbs1.0BTUlb°F(212°F32°F)Q_3 = m \cdot C_{\text{water}} \cdot \Delta T = 584 \, \text{lbs} \cdot 1.0 \, \frac{\text{BTU}}{\text{lb} \cdot °F} \cdot (212°F - 32°F) Q3=5841.0180=105120BTUQ_3 = 584 \cdot 1.0 \cdot 180 = 105120 \, \text{BTU}

4. Freezing Water to Ice at 32°F

Q4=mLatent Heat of Melting=584lbs144BTUlbQ_4 = m \cdot \text{Latent Heat of Melting} = 584 \, \text{lbs} \cdot 144 \, \frac{\text{BTU}}{\text{lb}} Q4=84096BTUQ_4 = 84096 \, \text{BTU}

5. Cooling Ice from 32°F to 25°F

Q5=mCiceΔT=584lbs0.5BTUlb°F(32°F25°F)Q_5 = m \cdot C_{\text{ice}} \cdot \Delta T = 584 \, \text{lbs} \cdot 0.5 \, \frac{\text{BTU}}{\text{lb} \cdot °F} \cdot (32°F - 25°F) Q5=5840.57=2044BTUQ_5 = 584 \cdot 0.5 \cdot 7 = 2044 \, \text{BTU}

Total BTUs Required

Total BTUs=Q1+Q2+Q3+Q4+Q5\text{Total BTUs} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 Total BTUs=22752+566480+105120+84096+2044=780492BTU\text{Total BTUs} = 22752 + 566480 + 105120 + 84096 + 2044 = 780492 \, \text{BTU}

Therefore, 780,492 BTUs are required to change 584 lbs of steam at 290°F to 584 lbs of ice at 25°F.

Would you like more details or have any questions?


Here are some related questions:

  1. How does latent heat impact phase changes?
  2. What is the significance of specific heat in temperature changes?
  3. Why is the latent heat of boiling so much higher than the latent heat of melting?
  4. How do pressure changes affect the boiling point of water?
  5. What role does energy conservation play in heat transfer problems?

Tip: Always double-check temperature differences and phase changes when dealing with heat transfer problems to ensure correct use of latent and specific heat values.

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Heat Transfer
Phase Changes

Formulas

Q = m * C * ΔT
Q = m * Latent Heat

Theorems

Law of Conservation of Energy
Specific Heat Capacity
Latent Heat of Melting and Boiling

Suitable Grade Level

College Level (Thermodynamics/Physics)