How many joules are absorbed when 200. g of water is raised from 10.0 c to 40.0 c ?

a. 126j
b. 840j
c. 25,080

option C is correct = 25,080 j

Explanation:

Data Given:

mass of water = 200 g

initial temperature = 10 °C

Final temperature = 40 °C

Amount of Heat = ?

Solution:

Formula used to calculate the amount of heat

             Q = c.m.ΔT . . . . . (1)

Where:

Q = amount of heat

c = specific heat of water = 4.18 J/g°C

m = mass

ΔT  = (t2 - t1) = Change in temperature

First we have to find ΔT

             ΔT  = (t2 - t1)  

             ΔT = (40 °C - 10°C )

             ΔT = 30 °C

Put values in above equation 1

            Q = 4.18 J/g°C x 200 g x 30 °C

            Q = 25080 j

So option C is correct