20920 joules of heat is absorbed by a pot of water with a mass of 500g in order to raise the temperature from 20°C to 30°C
Explanation:
mass of water=500g
final temperature=30°C
Initial temperature=20°C
change in temperature=30-20=10
specific heat of water=1 J/kg°C
[tex]energy required=mass of water \times temperature difference\times specific heat\\=500 \times10 \times \1\\=5000 calories\\=5kcal[/tex]
1 kcal=4184 joules
[tex]5 kcal=5 \times 4184=20920 joules[/tex]
The amount of heat absorbed by the pot of water is 20920 J
•Initial temperature (T₁) = 20 °C
•Final temperature (T₂) = 30 °C
•Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 30 – 20
ΔT = 10 °C
•Mass of water (M) = 500 g
•Change in temperature (ΔT) = 10 °C
•Specific heat capacity of water = 4.184 J/gºC
•Heat absorbed (Q) =?
Q = MCΔT
Q = 500 × 4.184 × 10
Q = 20920 J
Thus, the heat energy absorbed by the pot of water is 20920 J
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