What mass of water will change its temperature by 3.00 °C when 595 J of heat is added to it? The specific heat of water is 4.2 J/g °C

M = 36.5g

^ degrees sign

Q energy heat in JOULES J

m mass of the sample in GRAMS g

C specific heat in J/g^C

ΔT change in temperature ^C
or final temp- initial temp

Based on the information you've provided, we know the following variables:

ΔT= 3^C

Q = 525 J

C= 4.8 J
g×^C

All we have to do is rearrange the equation to solve for m. This can be accomplished by dividing both sides by C and ΔT to get m by itself like this:

m= Q
ΔT×C

Now, we just plug in the known values:

m= 525 J
3^C×4.8 J
g×^C

m=36.5g

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