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Sagot :
Answer:
B.
Solution given:
specific heat capacity of water [c]4.186 J/g °C
temperature[∆T]=?
mass[m]=3g
heat[Q]=8000J
we have
Q=mc∆T
8000=3*4.186*∆T
∆T=8000/12.558
∆T=637.04°C.
the temperature change is 637.04°C.
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