If 125 cal of heat is applied to a 60.0 g piece of copper at 20.0 degrees C, what will the final...

Question:

If {eq}125\ cal {/eq} of heat is applied to a {eq}60.0\ g {/eq} piece of copper at {eq}20.0 ^\circ C {/eq}, what will the final temperature be? The specific heat of copper is {eq}\rm 0.0920\ cal/(g \cdot ^\circ C) {/eq}. Express the final temperature numerically in degrees Celsius.

Heat Transfer:

The amount of heat transferred, q, to a material is reflected by its change in temperature, {eq}\displaystyle \Delta {/eq}T. The relationship between the quantities is expressed through the equation, {eq}\displaystyle q = mc\Delta T {/eq}, where m is the mass of the object and c is its specific heat.

Answer and Explanation:

Determine the final temperature of the metal sample using the equation, {eq}\displaystyle q = mc(T_f-T_i) {/eq}, wherein we are given q = 125 cal as the heat supplied, m = 60.0 g as the mass of the sample, c = 0.0920\ cal/g{eq}\displaystyle ^\circ {/eq}C as the specific heat and {eq}\displaystyle T_i {/eq} = 20.0{eq}\displaystyle ^\circ {/eq}C as the initial temperature. We proceed with the solution.

{eq}\begin{align} \displaystyle q &= mc\Delta T\\ \frac{q}{mc}&= T_f-T_i\\ \frac{q}{mc}+T_i &= T_f\\ \frac{125\ cal}{60.0\ g\times 0.0920\ cal/g ^\circ C}+ 20.0^\circ C &= T_f\\ \boxed{\rm 42.6^\circ C} &\approx T_f \end{align} {/eq}


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Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

Chapter 17 / Lesson 12
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