A small electric immersion heaters used to heat 71 g of water for a cup of instant coffee. The heater is marked “99 watts” (it converts electrical energy into heat energy at current rates). Calculate the time required to get all this water from 21 ° C to 100 ° C, without any regard to heat loss. (The specific heat of water is 4186 J / kg K. ·) please help??
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the main equation that you would be using is:
Q=mc(delta)T
Q is heat in joules
m is mass in kg
c is the specific heat of water (4186)
(delta)T is your difference in temp (100-21)
you solve for Q above and then use dimensional analysis to solve for the time (1 watt = 1 joule per second)
maybe around 4 mins..
heat(in kcal) = w*t/4200
Where w is the wattage ( here 99 watts), t is the time (here t in seconds) and heat is the energy required to raise the temperature of the water from a certain point to the other higher point ( here from 21 degrees to 100 degrees).
First of all,
Heat energy required to raise the temperature of 71 grams of water from 21 degree to 100 degree is
(100-21)*71 = 79*71 = 5609 cal = 5.609 kcal
now putting 5.609 in the above relation
5.609 = 99*t / 4200
= 23557.8 = 99t
=23557.8/99 = t
= 237.957 = t
Therefore time is 237.957 seconds.
Converting it into minutes.
237.957/60
= 3.96595 minutes
Converting 0.96595 into seconds
0.96595*60=57.95
Therefore total time will be 3 minutes and 58 seconds (approx.)