If 100000J of heat were added
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If 100000J of heat were added

[From: ] [author: ] [Date: 12-05-20] [Hit: ]
100 kJ = 4.∆T = 7.The water will warm up from 20º to about 28º-I do not understand what you mean by this : ..........
If 100000J of heat were added to 3000g of water at 20C instead of the unknown material, what will happen to water? Be specific as possible.

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U = c·m·∆T
U is energy in Joules to heat the material by ∆T
c is specific heat of the material in J/kg·K
m is mass in kg, ∆T is temperature change in ºC
specific heat of water is 4.186 kJ/kgC

100 kJ = 4.186 kJ/kgC x 3 kg x ∆T
∆T = 7.96º

The water will warm up from 20º to about 28º

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I do not understand what you mean by this : ".........instead of the unknown material, ......".
Somebody might read you better.
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