How much energy must be removed from a 500 g block of ice to cool it from 0 degree C to -40...

Question:

How much energy must be removed from a {eq}500 \ g {/eq} block of ice to cool it from {eq}0 ^\circ \ C {/eq} to {eq}-40 ^\circ \ C {/eq}? The specific heat of ice is {eq}2090 \ Jkg \cdot K. {/eq}

Heat Transfer:

The heat transferred into one substance is quantified with the equation, {eq}\displaystyle q = mc\Delta T {/eq}. It is proportional to the mass, m, the specific heat, c, and the change in temperature, {eq}\displaystyle \Delta {/eq}T, of the substance. This assumes that there are no physical nor chemical changes occurring in the substance.

Answer and Explanation:

Determine the energy, q, from the given condition using the equation, {eq}\displaystyle q = mc\Delta T {/eq}. For this problem, we are given a mass of ice of {eq}\displaystyle m = 500\ g = 0.5 \ kg {/eq}, a specific heat of {eq}\displaystyle c = 2090\ J/kg\cdot K = 2090\ J/kg\cdot ^\circ C {/eq}, and a temperature change of {eq}\displaystyle \Delta T = -40^\circ C - 0^\circ C = -40^\circ C {/eq}. We simply plug in the given values to determine the answer.

{eq}\begin{align} \displaystyle q &= mc\Delta T\\ &= 0.5\ kg \times 2090\ J/kg\cdot ^\circ C\times -40^\circ C\\ &=\boxed{\rm -41800\ J} \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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