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Find the amount of heat in joules required to raise the temperature of 28.0 grams of water from 30.0°C to 60.0°C.

User Helo
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1 Answer

1 vote

Answer: 3510 J

Step-by-step explanation:

q = m x Cs x delta T

m = 28.0 g

Cs = 4.18 J/g x ⁰C

delta T = Final temperature - Initial temperature = 60.0⁰C - 30.0⁰C = 30.0⁰C

q = m x Cs x delta T

q = 28.0 g x (4.18 J/g x ⁰C) x 30.0⁰C = 3511.2 J

3510 J for the correct number of significant figures.

Grams and ⁰C cancel and you are left with Joules.

User Mohamad Ghafourian
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