Question

How much heat energy is required to raise the temperature of 0.358kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of copper is 0.0920 cal/(g⋅∘C) .

Answer 1

this is physic hehe

Answer 2

lol i will try to solve it

Answer 3

is it 1.3105k cal?

Answer 4

\[C = E/(m \Delta T)\]
C = 0.092, m = 358g, delta T = 37
\[E =C(m \Delta T) = 0.092 * 358 * 37 = 1.2186 k \cal\]
i think i have copied the wrong number in last post

Answer 5

i guess that would be \[Q=mc \Delta t\]t
m=0.358kg=358g
c=0.0920 cal/g deg C
\[\Delta t\]=60-23=37 deg C
Q=358*0.0920*37=1218 cal

Answer 6

yeah 1218, or 1.218k cal

Answer 7

energy in cal or J

Answer 8

its just that you need to convert the mass of copper in g because rest all the values are in CGS system...

Answer 9

but is the ans correct?

Answer 10

sure

Answer 11

If 125 cal of heat is applied to a 60.0-g piece of copper at 20.0∘C , what will the final temperature be? The specific heat of copper is 0.0920 cal/(g⋅∘C) .

Answer 12

the same formula goes here
\[Q=mc Deltat\]
125=60*0.0920*(x-20)
125=5.52*(x-20)
22.65=x-20
x=42.65

Answer 13

correct?

Answer 14

yes yes...temp in C

Answer 15

ur welcum:)

Answer 16

hwy if you can help me with one more i would appreciat it

Answer 17

A cyclist rides at an average speed of 28mi/h .

Answer 18

If she wants to bike 181km , how long (in hours) must she ride?

Answer 19

hei can you help me

Answer 20

i got 4 hours is that right

Answer 21

speed=distance/time
use this formula...dat mi is miles right? i dnt know miles to km conversion. convert it and then just put the values in this formula.

Answer 22

convert km to miles...dat would be easy

Answer 23

i got 4 hours is it it right

Answer 24

28=112.47/t
t=112.47/28
t=4.01...u are correct:)

Answer 25

A classroom has a volume of 288m3

Answer 26

What is its volume in km3

Answer 27

i m not sure...ask in the physics group.

Answer 28

ok

Answer 29

Please post your questions in the appropriate group which is physics in this case.!