- #1
LostInScience
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Homework Statement
A 0.29 kg coffee mug is made from a material that has a specific heat capacity of 930 J/(kg·°C) and contains 0.23 kg of water. The cup and water are at 25°C. To make a cup of coffee, a small electric heater is immersed in the water and brings it to a boil in two minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.
Homework Equations
The heat required to heat something (Q)= specific heat * mass * change in temperature
1 watt = 1 joule/second
The Attempt at a Solution
so If I have to find watts used I need to find joules then divide that by 120 seconds because it takes 2 minutes to heat the water in the cup. However the problem says that the cup and the water always have the same temperature so wouldn't I have to combine the two equations to findthe heat required to warm them both to 100 degrees Celsius?
If so Q = (specific heat of cup + specific heat of water)(mass of cup + mass of water)(change in heat[temperature final-temperature initial])
So Q=(930+4186)(0.29+0.23)(100-25) Q=(5116)(0.52)(75) Q=199524 This is total joules required to heat the cup and water to 100 degrees Celsius from a temperature of 25 degrees Celsius. If I take total joules (199524) and divide by time (120 seconds) I will get joules per second as 1662.7 which is watts?
I've tried entering this into my homework and it's wrong. Where am I going wrong?
I tried the problem again using the equation Specific heat = Q/mass*change in temperature. I did this separately for the cup and water to get the joules required to heat each then added those two numbers together and divided by time.
For water I got 4186 = Q/(0.23)(75)=72208.5 joules
For the cup I got 930 = Q/(0.29)(75)=20227.5 joules
So total joules required = 92436/120 seconds = 770.3 watts?
Is that right?
Thanks for any help you can give me!