What is the heat in Joules required to raise the temperature of 100g H20 from 288K to 308K?
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: if Q is (-), heat is released and if (+), heat is absorbed
for water c = 4.184 J/g-K
substituting,
Q = 100*4.184*(308-288)
Q = 8368 J
hope this helps~ :)
8368J
To calculate the heat in Joules required to raise the temperature of a substance, you can use the equation:
Q = mcΔT,
where:
Q is the heat energy in Joules,
m is the mass of the substance in grams,
c is the specific heat capacity of the substance in J/g°C, and
ΔT is the change in temperature in degrees Celsius.
In this case, m = 100g, ΔT = 308K - 288K = 20K, and c is the specific heat capacity of water, which is approximately 4.18 J/g°C.
To calculate the heat, plug in the values:
Q = (100g) × (4.18 J/g°C) × (20K).
Simplify the equation:
Q = 8360 J.
Therefore, the heat required to raise the temperature of 100g of water from 288K to 308K is 8360 Joules.