Question

14. Heat energy equal to 25,000 J is applied to a 1200 g brick whose specific heat is 2.45 J/gºC.

Answer 1

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