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Heat

In an electrochemistry experiment the voltage between two electrodes is 10 Volts andthe current is 10 mA. Find the heat released after one minute.

For this I have calculated resistance R=V/I = 10/0.01 = 1000 ohms

and then used tV^2/R = 60*0.01^2 / 1000 = 0.000006

is this correct? :smile:
Original post by jacksonmeg
In an electrochemistry experiment the voltage between two electrodes is 10 Volts andthe current is 10 mA. Find the heat released after one minute.

For this I have calculated resistance R=V/I = 10/0.01 = 1000 ohms

and then used tV^2/R = 60*0.01^2 / 1000 = 0.000006

is this correct? :smile:
You need to post the full question if you want an answer. :smile:
Reply 2
Avatar for jacksonmeg
jacksonmeg
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Original post by uberteknik
You need to post the full question if you want an answer. :smile:


Umm.. this is the full question. In fact I copied and pasted it
Original post by jacksonmeg
Umm.. this is the full question. In fact I copied and pasted it
It's the word 'heat' which is ambiguous as it says nothing about the units the answer should be stated in. i.e. heat could be temperature difference or power or energy expended etc.

I'm assuming it's energy released in this example:

E=P x tE = P \mathrm{\ x \ } t

P=VIP = VI

E=VItE = VIt

E=10 x 10x10−3 x 60=6 JoulesE = 10 \mathrm{\ x \ } 10 \mathrm{x}10^{-3} \mathrm{\ x \ }60 = 6 \mathrm{\ Joules}

Alternately:

R=VI=100.01=1000ΩR = \frac{V}{I} = \frac{10}{0.01} = 1000 \Omega

P=V2R=1001000=0.1 WattsP = \frac{V^2}{R} = \frac{100}{1000} = 0.1 \mathrm{\ Watts}

E=P x t=0.1 x 60=6 JoulesE = P \mathrm{\ x \ }t = 0.1 \mathrm{\ x \ }60 = 6 \mathrm{\ Joules}
Reply 4
Avatar for jacksonmeg
jacksonmeg
OP
11
Original post by uberteknik
It's the word 'heat' which is ambiguous as it says nothing about the units the answer should be stated in. i.e. heat could be temperature difference or power or energy expended etc.

I'm assuming it's energy released in this example:

E=P x tE = P \mathrm{\ x \ } t

P=VIP = VI

E=VItE = VIt

E=10 x 10x10−3 x 60=6 JoulesE = 10 \mathrm{\ x \ } 10 \mathrm{x}10^{-3} \mathrm{\ x \ }60 = 6 \mathrm{\ Joules}

Alternately:

R=VI=100.01=1000ΩR = \frac{V}{I} = \frac{10}{0.01} = 1000 \Omega

P=V2R=1001000=0.1 WattsP = \frac{V^2}{R} = \frac{100}{1000} = 0.1 \mathrm{\ Watts}

E=P x t=0.1 x 60=6 JoulesE = P \mathrm{\ x \ }t = 0.1 \mathrm{\ x \ }60 = 6 \mathrm{\ Joules}

Ahh I see what you mean, thanks for the response !

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