m = 1000 grams
c = 0.129 J/g°C
ΔT = (Tfinal – Tinitial) = (75 °C – 25 °C) = 50 °C
Plug these values into the specific heat equation from above.
Q = mcΔT
Q = ( 1000 grams)·(0.129 J/g°C)·(50 °C)
Q = 6450 J
Answer: It took 6450 Joules of energy to heat the lead cube from 25 °C to 75 °C.
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Answers & Comments
m = 1000 grams
c = 0.129 J/g°C
ΔT = (Tfinal – Tinitial) = (75 °C – 25 °C) = 50 °C
Plug these values into the specific heat equation from above.
Q = mcΔT
Q = ( 1000 grams)·(0.129 J/g°C)·(50 °C)
Q = 6450 J
Answer: It took 6450 Joules of energy to heat the lead cube from 25 °C to 75 °C.