955 grams of water at 25.0 C is heated to 85.5 C. Calculate the amount for Q in Joules?

Question

a.57800 J

b. 242000 J

c. -57800 J

d. -242000 J

Answer 1

Hello, Kelly,

What is the specific heat of water that is provided with the problem. I'll use the standard 4.186 kJ/gC, since the units are given in grams and C. Please note that this is kJ (kiloJoules).

Use the equation q = cmDt, where q is heat/energy, c is the specific heat of water, m is the mass, and Dt is the change in temperature.

q = (4.186kJ/gC)*955g*(85.5C - 25.0C) [Update: As per comment from J.R. S. below, the specific heat of water is 4.186kJ/kgC, not g as I used in the equation. That means the equation should read: q = (4.186kJ/kgC)*0.955kg*(85.5C - 25.0C). Note that the 955g must be expressed as kg, so I used 0.955kg in the equation]

The kilograms and C cancel, leaving kJ.

I see an answer that numerically matches what I calculate. [An answer now matches]

Bob