Question

1,500,000 J of energy is applied

to 20kg of Water at 280k . What is the
new Temperature ? how long will a 350 watt
Energs Source require to raise T ?

Answer 1

a) T = 298 K

b) t = 4285.71 s

Step-by-step explanation:

• Q = mCpΔT
• 1 watt ≡ 1 (J)/(s)

∴ Q = 1500000 J

∴ m H2O = 20 Kg = 20000 g

∴ Cp H2O = 4.184 J/g*K

∴ T1 = 280 K

∴ ΔT = T2 - T1

a) T2 = ?

⇒ ΔT = Q/m*Cp

⇒ ΔT = (1500000 J )/(20000 g)(4.184 J/g.K)

⇒ ΔT = 17.925 K

⇒ T2 - T1 = 17.925 K

⇒ T2 = 280 K + 17.925 K

⇒ T2 = 297.925 K ≅ 298 K

b) Q = 350 watt ⇒ t = ?

∴ Q = 350 J/s = m*Cp*ΔT

⇒ t(s) = Q/m*Cp*ΔT

⇒ 350 J/s = (20000 g)*(4.184 J/g.K)*(298 K - 280 K) = 1500000 J

⇒ 1/t = (350 J/s)/(1500000 J)

⇒ 1/t = 2.33 E-4 s-1

⇒ t = 4285.71 s