How many joules are needed to warm 45.0 grams of water from 30.0 degrees C to 75.0 degrees C?

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 (final-initial 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

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Alleei Alleei · Answer 2 · 2020-03-07T01:52:50-05:00

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 (75.0-30.0)^oC[/tex]

[tex]q=8464.5J[/tex]

Thus, the amount of heat needed are, 8464.5 J