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The temperature of a 12.58 g sample of calcium carbonate [CaCO3(s)] increases from 23.6 Celsius to 38.3 Celsius. If the specific heat of CaCO3 is 0.82 J/g-K, how many joules of heat are absorbed during this process

User Dasper
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1 Answer

6 votes

Answer:

THE AMOUNT OF HEAT ABSORBED DURING THE REACTION IS 151.64 JOULES

Step-by-step explanation:

Mass = 12.58 g

Specific heat of CaCO3 = 0.82 J/g K

Change in temperature = 38.3 C - 23.6 C = 14.7 C

How many joules of heat are absorbed?

Heat of a reaction is the amount of heat absorbed or evolved when a gram of a sample is raised by 1 C.

Heat = mass * specific heat * change in temperature

Heat = 12.58 * 0.82 * 14.7

Heat = 151.64 J

The amount of joules absorbed during this reaction is 151.64 J.

User Volomike
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