the final temperature after 840 Joules is absorbed by 10.0g of water at 25.0oC at 45°C.
Explanation:
Data given:
initial temperature (T)= 25 degrees Celsius
mass of water = 10 gram
cp of water = 4.2J/gram Celsius
energy absorbed (q) = 840 joules
change in temperature = final temperature(t2) - initial temperature (t1)
so ΔT ( t-25)
Using the equation.
q= mcΔT
putting the values in the equation:
840 = 10 x 4.2 x (t-25)
t -25 = [tex]\frac{840}{10 x 4.2}[/tex]
t = [tex]\frac{840}{10 x 4.2}[/tex] +25
= 45 °C
The final temperature of 10 gram of water which absorbed 840 joules of energy at an initial temperature of 25 degrees and final temperature of 45 degrees.
