Question

1.Calculate the number of joules released when 72.5 grams of water at 95.0 degrees Celsius cools to a final temperature of 28.0 degrees Celsius. i am getting ready for a test and do not understand how to calculate these types of problems. this is an example from the lesson i know the answer but i would like to know how to get there.

Answer 1

use the equation for calorimetry:\[Q = m*C*\Delta T\]where Q is the heat in joules, m is the mass of the system (in grams), C is the specific heat of the system in J/(g*C), and \(\Delta T\) is the change in temp of the system. You've got 3 of the 4 pieces, if you remember the C of water is always 4.18J/gC

Answer 2

do you understand it? @Avenged7x

Answer 3

whati don't get is after you put the equation like this: q= 72.5 x 4.18 J / g x C x .28 - .95 where do I from there?

Answer 4

do the math and solve for Q

Answer 5

Exactly how it is written correct?

Answer 6

i didn't get what u just said
anyways look
temperature was 95 and it dropped to 28
that means 95- 28= 67 is your change in temperature
so
\[Q= mc \Delta T\]
\[Q= 72.5\times 4.18 \times 67\]
and get the answer
notice: the temperature Dropped means energy Released thus sign of your final answer will be negative ' - '

Answer 7

your change in temp is written incorrectly, the subtraction has to occur first, before the multiplication\[72.5g * 4.18\frac{J}{g*C}*(28C - 95C)\]

Answer 8

oooooh ok that's different than my teacher explained so my answer is -20164.3?

Answer 9

I had the answer but I closed out my class on accident

Answer 10

yes :)

Answer 11

wait how r u getting -20164.3?

Answer 12

its gonna be 20304.35 J