What is the final temperature after 840 Joules is absorbed by 10.0g of water at 25.0oC?

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 =

t = +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.