How much heat is needed to raise 100g of water from 34°C to 100°C? The specific heat of water is 4.18 J/g∙C?
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Answer:
T1=34°C = 273+34= 307
T2=100°C = 273+100=373
∆T= T2_T1
=373_307
=66
c=4.18
∆Q= ?
m=100 grams= 0.1kg
∆Q= mc∆T
∆Q=0.1×4.18×66
∆Q=27.588
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