Question

The specific heat of diamond is 0.127 space straight J times straight g to the power of negative 1 end exponent times degree straight C to the power of negative 1 end exponent. If a 2 carat diamond weighing 400.0 milligrams at room temperature (22.0 °C) absorbed 5.55 J of heat, what is the final temperature of the diamond? 

          The specific heat of diamond is 0.127 space straight J times straight g to the power of negative 1 end exponent times degree straight C to the power of negative 1 end exponent. If a 2 carat diamond weighing 400.0 milligrams at room temperature (22.0 °C) absorbed 5.55 J of heat, what is the final temperature of the diamond?
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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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The specific heat of diamond is 0.127 space straight J times straight g to the power of negative 1 end exponent times degree straight C to the power of negative 1 end exponent. If a 2 carat diamond weighing 400.0 milligrams at room temperature (22.0 °C) absorbed 5.55 J of heat, what is the final temperature of the diamond?
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The specific heat of most solids is nearly constant over a wide temperature range. Not so for diamond. Between 200 K and 600 K, the specific heat of diamond is reasonably well described by c = 2.8T - 350 J/kg · K, where T is in K. For gemstone diamonds, 1 carat = 200 mg. Part A How much heat energy is needed to raise the temperature of a 4.0 carat diamond from -50 °C to 250 °C? Express your answer with the appropriate units. Q = 153.2 J

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Transcript

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00:01 Hello students, we will discuss a problem from thermochemistry and we need to discuss or calculate the amount of the heat that is supposed to be required to raise the temperature of diamond from 36 degrees centigrade to 52 degrees centigrade and mass of the diamond is given as 0 .702 gram a specific heat of the diamond is also given that is 0 .5050 jule gram per degree centigrade in order to get the answer for this particular question we need to we need one equation the formula and the formula is q is equal to m multiplied by c multiplied by delta t where we know that m is equal to mass and this mass should be into grams c is equal to a specific heat and delta t is the raise in the temperature and and here it is 36 degrees centigrade to 52 degree centigrade.
01:13 And the values are given as we know masses 0 .702 gram.
01:20 Specific heat is equal to 0 .50.
01:28 Jule per gram per degree centigrade...
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