Question

42.5 grams of an unknown substance is heated to 105.0 degrees Celsius and then placed into a calorimeter containing 110.0 grams of water at 24.2 degrees Celsius. If the final temperature reached in the calorimeter is 32.4 degrees Celsius, what is the specific heat of the unknown substance?

Show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 J/(°C x g).

Answer 1

@JoannaBlackwelder is this where that equation i quoted earlier comes it?

Answer 2

Yup!

Answer 3

what was the first part of the equation i know its Q=. x C x deltaT

Answer 4

m

Answer 5

i keep getting off task cause im at work while doing this lol but how do i begin this... im clueless at this junk cause its too close to algebra lol

Answer 6

Aw. Yeah, it is a lot of algebra.

Answer 7

The idea is to find the amount of heat that the water absorbs.

Answer 8

Since the calorimeter is isolated, all of the heat that got absorbed by the water came from the unknown substance.

Answer 9

Can you find the Q for water?

Answer 10

what does q stand for?

Answer 11

Heat or energy

Answer 12

nvm its the heat gained or lost by the system isnt it?

Answer 13

Yup :)

Answer 14

8.2?

Answer 15

How did you get that?

Answer 16

the change in the equation is a gain of 8.2

24.2+8.2 = 32.4

Answer 17

The change in temp is 8.2, but that doesn't finish solving for Q.

Answer 18

I've gotta go. Be back on later.

Answer 19

ill be here