Question

What is the temperature change for a substance if 198 g releases 0.234 kJ of heat and has a specific heat of 0.360 J/g°C? And why?

Answer 1

use the equation: \(q=m*C_p*\Delta T\)

Answer 2

it says the answer is 3.28 how do I get that answer

Answer 3

plug your variables into the equation

Answer 4

can you lay it out for me with the variables plugged in

Answer 5

nope, cuz then i'd be doing it for you and you wouldn't learn anything.
you should try it, i'll help you correct it

Answer 6

all I keep getting is 16.67

Answer 7

show me what you've done and i'll help you correct it

Answer 8

I did 198*0.234*.360

Answer 9

you're looking for the change in temperature, \(\Delta T\).

do you know what the other variables in this "\(q=m*C_p*\Delta T\)" are?

Answer 10

I know m is 198 and Cp is 0.360 so does that mean q is .234?

Answer 11

exactly. do you know what the sign of q (positive or negative) should be when the system is releasing energy?

Answer 12

I do not, I am sorry I know this must be frustrating but I was not supposed to be in Chemistry in the first place, let alone honors

Answer 13

no worries, it's pretty easy if you think about it. It always helps to know the definitions of everything.

So when a system is releasing energy, it's called an exothermic process, q is negative. Conversely, when a system is absorbing energy, it's called an endothermic process, q is positive.

Answer 14

can you plug the variables into the equation with the correct signs now?

Answer 15

-0.234=198*0.360* ΔT

Answer 16

yep that's good. except 0.234 kJ and 0.360 J/g*C need to be in the same units.

1 kJ = 1000 J

so 0.234 kJ * 1000J/kJ= 234 J

-234=198*0.360* ΔT

Answer 17

this will give you the right answer.

Answer 18

I am still not getting it, how do I solve with the -.234 on the left side

Answer 19

remember it's not -0.234 kJ it's -234.0 J, units need to be the same.

solve it like any algebraic equation:

-234=198*0.360* ΔT

isolate ΔT=\(\dfrac{-234\;J}{198\;g*0.360\;J/g*^oC}\)

Answer 20

thank you so much that really really helped

Answer 21

no problem ! i hope you understood it well enough to do it by yourself next time