Question
A 0.250-kg block of a pure material is heated from 20.0C20.0^\circ\textrm{C} to 65.0C65.0^\circ\textrm{C} by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed.
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Final Answer

387 J/kgC387\textrm{ J/kg}\cdot\textrm{C}^\circ
This is copper.

Solution video

OpenStax College Physics, Chapter 14, Problem 6 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. 0.250 kilograms of material is heated from 20.0 degrees Celsius to 65.0 degrees Celsius and it takes 4.35 kilojoules to make that temperature change happen. We'll write our kilojoules in units of joules by multiplying by 10 to the 3 since that's what the prefix 'kilo' means and we have a formula that the heat added is the mass times its specific heat of the substance times the change in temperature and our job here is to find out what c is and then use that to figure out what material we are dealing with in this table of specific heats. So ΔT is T final minus T initial; we are gonna divide both sides by m and divide by T f minus T i and then we'll solve for c and we switch the sides around so we have the unknown on the left so c equals Q divided by m times the change in temperature. So that's 4.35 times 10 to the 3 joules divided by 0.250 kilograms times 65.0 degrees Celsius minus 20.0 degrees Celsius and that is 387 joules per kilogram per Celsius degree. And if we look up this value in the "Specific Heat" table, this column here has units of joules per kilogram per Celsius degree and we find 387 here and it is 'Copper.' So this material is copper.