if 75 g of liquid water (C=4.18 J/g-°C) in a calorimeter changes temperature from 25°C to 75°C, how much heat was transferred? use the equation C=q/mΔT

Question

aaronq · Answer

Use the equation provided

anonymous · Answer

can you walk me through it maybe? thats why i asked on here..... @aaronq

aaronq · Answer

I can help you correct any mistakes you make if you attempt it

anonymous · Answer

well its gonna be a long time because i dont know how to do it. thats why im asking for help.

aaronq · Answer

It's rather simple, assuming you know what the variables are. Plug in the variables where they go and solve.

I don't mind helping you, hence why i'm here, but you have to put in some effort.

anonymous · Answer

assuming i know the variables. i dont. im not asking you to do it for me. but if you could idk start me off or something it would help alot

aaronq · Answer

Sure thing, the variables are the following:

q = heat
m= mass
C = specific heat capacity (given in the question)
$\sf \Delta T$ is the change in temperature, explicitly, $\sf  \Delta T=T_{final}-T_{initial}$

anonymous · Answer

the capacity is gonna be what 75? right?

aaronq · Answer

no, the specific heat capacity for water is: C=4.18 J/g-°C

aaronq · Answer

75 g is the mass

anonymous · Answer

i meant the other 75. but okay

aaronq · Answer

That's the temperature

anonymous · Answer

okay. writing out the equation

anonymous · Answer

i dont know

aaronq · Answer

lol you're so close. 

$q= m*C*\Delta T$

Just plug the stuff in and solve, try it here

anonymous · Answer

i dont know what m or T is

anonymous · Answer

and that equation is different....

aaronq · Answer

m stands for mass, the mass of the water in this case

$\Delta T$ is the difference in temperature.

Whats' the difference in temperature between 15 $^oC$ and 25 $^oC$?

$\Delta T=25^oC-15^oC=10^oC$

aaronq · Answer

The equation is the exact same, it's just rearranged with algebra

anonymous · Answer

well im not passing algebra soooo i was confused. sorry

aaronq · Answer

No need to apologize. I think algebra is necessary to solve these problems, a pre-requisite if you may. you're going to run into a lot of of these kinds of problems where you have to use algebra.

anonymous · Answer

15.7kJ