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When 25.0 grams of water are cooled from 20.0 degrees Celsius to 10.0 degrees Celsius the number of joules of heat energy released is?

User Stmatengss
by
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2 Answers

2 votes

Answer: The amount of heat released is -1046 J

Step-by-step explanation:

To calculate the amount of heat released, we use the equation:


q=mc\Delta T

where,

q = heat released = ?

m = mass of water = 25.0 g

c = specific heat capacity of water = 4.184 J/g.°C


\Delta T = change in temperature =
T_2-T_1=(10.0-20.0)^oC=-10.0^oC

Putting values in above equation, we get:


q=25.0g* 4.184J/g.^oC* (-10.0^oC)\\\\q=-1046J

Hence, the amount of heat released is -1046 J

User Justin Bozonier
by
8.7k points
5 votes

Heat energy released here

Q = mass x specific heat capacity of water x deltaT


= 25.0 x 4.184 x (-10.0)


= - 1046 Joules


1046 Joules of heat energy is released in this process.




User Mare Infinitus
by
8.6k points

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