14. Heat energy equal to 25,000 J is applied to a 1200 g brick whose specific heat is 2.45 J/gºC.
1 answer:
Answer:
∆T = 51 ºC
T2 = 76 ºC
Explanation:
(a)
∆Q = 25,000 J
m = 1200 g
c = 2.45 J/gºC
∆T = ?
∆Q = mc∆T
∆T = ∆Q/mc
∆T = 25000/(1200×2.45)
∆T = 51 ºC
(b) If T1 = 25 C
∆T = T2 - T1
T2 = ∆T + T1
T2 = 51+25
T2 = 76 ºC
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