Question

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?

Answer 1

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.  

     

Answer 2

Answer: The amount of heat released is -1046 J

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:

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

Hence, the amount of heat released is -1046 J