Question

sooo beyond stumped!! please please help

A piece of an unknown substance weighing 124.0 grams is heated in boiling water to 100.0oC. When the substance is placed in a calorimeter containing 300.0 grams of water, the temperature of the water increased from 20.0oC to 26.0oC. What is the specific heat of the unknown substance, assuming all the heat is transferred to the water?

Answer 1

do you recall the heat equations?

Answer 2

from what I remember the formula for sh is SH = q/m*change in temp

Answer 3

that sounds about right to me ...

Answer 4

Q(cold water) + Q(unknown substance) = 0 si what rings a bell for me

Answer 5

hmmm...I think q...for this equation anyway is equal to joules...which should be 4.184J
But I'm not sure.

Answer 6

124.0 grams, 100.0oC to 26 right?

300.0 grams of water, 20.0oC to 26.0oC, 4.184 (water Cp)

Answer 7

-100
26
----
- 74 change in temp of substace

and a +6 change in temp of the water

Answer 8

if so, i want to say this is the runup

300(4.184)(6) + 124(-74)(Cp) = 0

124(-74)(Cp) = -300(4.184)(6)

Cp = -300(4.184)(6)/[124(-74)]

Answer 9

ok...well I'll try that and se how it goes. Thanks for the help.

Answer 10

youre welcome ... i was looking over the "lab report" i did for my own calorimeter project