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Heat is added to a silver bar of mass 0.034 kg and length 0.28 m. The bar expands by an amount of...

Question:

Heat is added to a silver bar of mass 0.034 kg and length 0.28 m. The bar expands by an amount of {eq}3.8 \times 10^{-4} {/eq} m. How much heat was added?

Thermal Expansion:

When the heat is supplied to the metal, then its length is expanded, it is known as thermal expansion. The change in the length is directly proportional to the change in the temperature.

Answer and Explanation:

Given data

Mass of the bar {eq}(m) = 0.034 \ kg {/eq}

Length of the bar {eq}(L) = 0.28 \ m {/eq}

Change in the length of the bar {eq}(\Delta L) = 3.8 \times 10^{-4} \ m {/eq}

Now, the change in the length is given by

{eq}\Delta L = \alpha \times L \Delta T \\ 3.8\times 10^{-4} = (18 \times 10^{-6}) (0.28) \Delta T \\ \Delta T = 75.397^\circ C {/eq}

Where

  • {eq}\alpha = 18 \times 10^{-6} \ /^\circ C {/eq} is the coefficient of the thermal expansion

Now, the heat added to the bar would be

{eq}Q = mC \Delta T \\ Q = (34 \ g) \times (0.24 \ J/g C) \times (75.397^\circ C) \\ Q = 615.24 \ J {/eq}

where

  • {eq}C = 0.24 \ J/g C {/eq} is the specific heat of the silver

Learn more about this topic:

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Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

Chapter 17 / Lesson 12
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