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: ΔQ = mcΔT : ΔQ = (60g) (0.385 J g−1 K−1) (60 °C same as 60 Kelvin) : :: = 1386 Joules

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How much heat is absorbed by 60.0 grams of copper when it is heated from 293 Kelvin to 353 kelvin?

The specific heat capacity of copper is 0.385 J/g°C. You can use the formula Q = mcΔT, where Q is the heat absorbed, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values, you can calculate the heat absorbed by the copper.


How much heat is absorbed by 1.0 g of iron when its heated to 15 degrees celsius?

The specific heat capacity of iron is 0.449 J/g°C. To calculate the heat absorbed, use the formula Q = mcΔT, where Q is the heat absorbed, m is the mass in grams, c is the specific heat capacity, and ΔT is the temperature change. Plugging in the values, the heat absorbed by 1.0 g of iron heated to 15°C is 6.735 J.


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How much heat must be absorbed by 50 grams?

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Which is a better conductor of heat air or a copper wire?

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A 38 kg block of lead is heated from -26C to 180C how much heat energy does the block of lead absorb to increase his temperature this much?

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Why does copper sulfate change mass when heated?

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How much heat is absorbed by 60.00 grams of copper when it is heated from 293 Kelvin to 353 Kelvin?

1,386J the formula needed in q=mC Delta T Delta is usually written as a triangle, but that not available here. q is heat m is mass C is specific heat (of copper in this case) every substance has a constant delta T means change of temp, in degrees centigrade to match units of "C" so, calculate delta T, convert to centigrade first then plug in numbers, multiply and go. 353 K - 293K = 80 Centigrade - 20 Centigrade = 60 Centigrade see... q = (60.00 G)(0.39J/gxdegrees centigrade)(60 C) all units cancel out except joules, the unit of heat, 1404 Joules 1386!!


How hot is gold when heated?

It depends how much heat is applied and the mass of the gold.


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When 30 grams of water is converted into steam how much heat is absorbed?

q = mHvq = heatm = mass (30g)Hv = heat of vaporization (2,260J/g)q = (30g)(2,260J/g)q = 67,800JWhen 30 grams of water is converted into steam, how much heat is absorbed?67,800J of heat, also represented as 67.8kJ of heat is absorbed.