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

@ganeshie8

Answer 2

Use \[Q=m*c*\Delta(T)\]
We heated Water using the unknown susbtance
C=specific heat capacity of water (4.18 J g-1 °C-1),
m = mass of water heated (kg),
DeltaT = change in temperature of the water (°C)
Q will represent the heat transferred from the 42.5 g of the unknown substance to Water.

Answer 3

a rise in temperature is realized to Water but a decrease in temperature of unknown Substance.

Answer 4

I dont understand

Answer 5

When the heated 42.5 substance was placed heat energy was transferred to water and lead to a rise in the water's temperature. we calculate the quantity of heat transferred using the equation above

Answer 6

How do i do the work

Answer 7

plug in your givens mass of water=110 specific heat capcity of water is 4.18 and Delta T=(final Temp-Initial Temp)=(32.4-24.2))
q=mcDeltaT

Answer 8

im still confused :/

Answer 9

about?

Answer 10

how to get everything and how to get it

Answer 11

how to do it*

Answer 12

When you heat water what happens to its temperature?

Answer 13

it rises

Answer 14

Great, in your question its the same thing a heated substance is placed in water.
Before the heated susbtance was added water had an intial susbtance of 24.2. As the heated substance is placed temperature increases from 24.2 to 32.4

Answer 15

initial temperature of 24.2 **

Answer 16

so how will we write this all out as a equation

Answer 17

Specific heat capacity
the heat required to raise the temperature of the unit mass of a given substance by a given amount. The value is given in your question which is 4.18 J/(°C x g).
Q=110*4.18*(32.4-24.2)

Answer 18

so it would be 4.18 J/(°C x g). Q=110*4.18*(32.4-24.2)

Answer 19

yes that's the quantity of heat transferred from the 42.5 grams of unknown substance to water

Answer 20

So would that be the answer to the whole thing?

Answer 21

no, what happened to the temperature of the unknown substance when it was placed in water?

Answer 22

it rose

Answer 23

the unknown substance lost some of its heat energy and was given to water so its temperature drops

Answer 24

ok i still dont know what the answer is

Answer 25

what was the final temperature recorded in calorimeter?

Answer 26

32.4

Answer 27

and what was the starting temperature of the 42.5 grams of unknown substance?

Answer 28

105.0

Answer 29

heat loss=heat gain
so 42.5*C*(32.4-105)=110*4.18*(32.4-24.2)

Answer 30

so thats the answer?

Answer 31

-3085.5*C=3770.36
C=3770.36/-3085=1.22

Answer 32

so how do we get that answer without explaining it?

Answer 33

its 1.22

Answer 34

-3085.5*C=3770.36 C=3770.36/-3085=1.22 so that is the whole process of getting that answer

Answer 35

we divided the 3370.36(which represent the heat gained)
no the part above too! without the above part of calculation you couldnt use heat lost=heat gained because you wouldnt have know the amount of heat transferred.

Answer 36

so what is the whole calculation?

Answer 37

Q=mc*Delta T
Q=110*4.18*(32.4-24.2)
Heat lost=Heat gained
mc*DeltaT=3770.36
42.5*c*(32.4-105)=3770.36
solve for c

Answer 38

c is 1.22

Answer 39

yes

Answer 40

Where did c come from? I thought is was this,: (4.18 J g-1 °C-1), so why would we have to solve for c?

Answer 41

Wait a second....wouldnt C actually be 4.18 J/(° C × g)....