A calorimeter containing 100 mL of water is calibrated by
passing a 3.00 A current through the instrument for 36.0 s
at a potential difference of 3.50 V. The temperature rises by
0.82°C.
Potassium hydroxide weighing 0.654 g is added to the
calorimeter and dissolved rapidly by stirring. The temperature
rises from 20.82°C to 22.23°C.
Determine the delta H

i know the calibration factor is 461J/celcius

i got -55.8 kj/mol but the answers say -55.8MJ/mol

did i get it wrong?

Question

A calorimeter containing 100 mL of water is calibrated by
passing a 3.00 A current through the instrument for 36.0 s
at a potential difference of 3.50 V. The temperature rises by
0.82°C.
Potassium hydroxide weighing 0.654 g is added to the
calorimeter and dissolved rapidly by stirring. The temperature
rises from 20.82°C to 22.23°C.
Determine the delta H

i know the calibration factor is 461J/celcius

i got -55.8 kj/mol but the answers say -55.8MJ/mol

did i get it wrong?

koikkara · Answer

$\large\color{darkgreen}{~''Nice~To~Meet~You~!!"}$ @bestie

koikkara · Answer

$\star~Formula:$ 
(Δ H) = mcΔT 
m= mass 
c= specific heat (constant depending on substance) 
Δ T = temperature Change

$Can~You~Solve~Now~??$ @bestie

anonymous · Answer

hello :) yes i used that, but i got a different ans from the book?

koikkara · Answer

Alright, can You Please provide your $Calculation~Steps$ here, Too..!!

anonymous · Answer

i have attached them as a pic :)

koikkara · Answer

Hmm... You got value in "J\mol" , then to have value in 'Kj/mol'.... convert it !!

1000 J/mol is = 1 kJ/mol.
55758 j\Mol/1000= 55.8 Kj/mol

Does this helps ?

anonymous · Answer

thats what i got, but the ans says MJ/mol?

koikkara · Answer

Alright, what will u get when u convert KJ/mol $\to$ MJj/mol ??

koikkara · Answer

I tried, but it doesnot come even close to 55.8Kj/Mol, that means, the provided unit is wrong.. and the answer You got is Right !!

koikkara · Answer

$Hope~It~helps~!!$

anonymous · Answer

thank you