Question

What mass of liquid water a t room temperature (25°C) can be raised to its boiling point with the addition of 40.0 kJ of energy?

Answer 1

to get the heat released or absorbed,
Q = mc(T2-T1)
where
m = mass of substance (units in g)
c = specific heat capacity (units in J/g-K)
T2 = final temperature
T1 = initial temperature
*note: units of temp doesn't matter if it's in degree Celsius or Kelvin, as long as T1 and T2 have the same units (since it's difference in temperature or delta,T)
**note: if Q is (-), heat is released and if (+), heat is absorbed

for water, c = 4.184 J/g-K
substituting,
40 * 10^3 = m*(4.184)*(100 - 25)
40000/(4.184)(75) = m
m = 127.47 g

hope this helps~ :)

Answer 2

Jai why did you multiply 40 by 10^3

Answer 3

Converting it to joules

Answer 4

Since it is 40 kilojoules; the ratio for kilojoules to joules is 1 to 1000, in other words by multiplying 40 by 10^3 (.0001), it is the same as dividing by 1000 to convert the kilojoules to joules.