Question

Metal: Aluminum
Mass of metal: 27.776 g
Volume of water in the calorimeter: 26.0 mL
Initial temperature of water in calorimeter: 25.3 °C
Temperature of hot water and metal in hot water bath: 100.5 °C
Final temperature reached in the calorimeter: 31.6 °C


1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.

The water has absorbed the heat of the metal. So, qwater = qmetal
q = m x c x delta t

Answer 1

How much cool water and how much hot water did you have?

Answer 2

Actually, it doesn't matter because you're only calculating the change in heat of the surroundings.

just use the formula: q=m*C*(Tf-Ti)

Answer 3

i dont think it matters

Answer 4

i just need help setting up the equation

Answer 5

(m) Volume of water in the calorimeter: 26.0 mL -> 1mL=1g (b/c density is 1g/mL)
(Ti) Initial temperature of water in calorimeter: 25.3 °C
(Tf) Final temperature reached in the calorimeter: 31.6 °C
(C) specific heat capacity of water is 4.18 J / (g × °C)

Answer 6

thats all the info you need, the rest (like the mass of the metal) is to confuse you

Answer 7

ok thank you so much this really helps

Answer 8

no problem

Answer 9

Using the formula qmetal = m × c × ΔT, calculate the specific heat of the metal.
^ what about this?

Answer 10

you know that q(water=)-q(metal) because the assumption is made that no heat was lost to other surroundings (like the room, your hands, etc)

use the q you found in the previous part and the change in temperature of the metal and it's mass.

q=m*C*(Tf-Ti)

Answer 11

so that whole first part i got 685, i just dont know what to do with the whole q(water=)-q(metal)

Answer 12

q(metal) is positive because heat was lost, q(water) is negative because heat was gained.

so for q(water)=685 , then q(metal)=-685

Answer 13

sorry i screwed up the brackets, it should read: q(water)=-q(metal)

Answer 14

so that is the specific heat?

Answer 15

nope, you gotta solve the equation using the data of the metal.

Answer 16

specific heat capacity is basically a measure of how much a certain substance retains heat.

Answer 17

so is it something like: 685
-------
26 x 6.3

Answer 18

you have the concept right, but the change in temperature of the metal is not the same as the change in temp of the water.

Answer 19

and you're using the mass of the metal not the mass of the water

Answer 20

ohh so 27.776
---------
26 x 6.3

Answer 21

not quite, write in the units, it's easier to see where you went wrong.

q=m*C*dT -> solve for C -> C=q/(m*dT)

btw, dT=Tf-Ti

Answer 22

but im not using C in specific heat am i?

Answer 23

C of the metal is what the question is asking you to find

Answer 24

lol oh yeeeah ok so hang on lemme try to figure this out

Answer 25

hah okay

Answer 26

685
-----------
27.776 x 6.3

Answer 27

it's right, except the change in temperature of the metal, it went from 100.5 to 31.6, clearly more than 6.3

Answer 28

OMG I I ACTUALLY GET IT!!!

Answer 29

so after i do this we're done?

Answer 30

yes, but make sure you include the units at the end

Answer 31

what would be the unit for specific heat??

Answer 32

when you include the units in the equation you'll see because they won't cancel out

Answer 33

oh ok thanks!!!

Answer 34

no problem dude !