How many joules are needed to warm 45.0 grams of water from 30.0 degrees C to 75.0 degrees C?
Chemistry
Angelica16
Question
How many joules are needed to warm 45.0 grams of water from 30.0 degrees C to 75.0 degrees C?
2 Answer

1. User Answers sissyrose
q=m x Cp x ∆T
m is the mass of the water, Cp is the specific heat of water and ∆T is the change in temperature of the water (finalinitial temperature). q is the energy involved in the reaction, measured in joules.
q=(45.0) x (4.184 Jg^1/°C^1) x (45°C)
q=8472.6 Joules 
2. User Answers Alleei
Answer : The amount of heat needed are, 8464.5 J
Explanation :
Formula used :
[tex]q=m\times c\times (T_{final}T_{initial})[/tex]
where,
q = heat gained = ?
m = mass of water = 45.0 g
c = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]75.0^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]30.0^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=45.0g\times 4.18J/g^oC\times (75.030.0)^oC[/tex]
[tex]q=8464.5J[/tex]
Thus, the amount of heat needed are, 8464.5 J