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An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at 20°C. A block of iron of mass 0.2 kg at 100°C is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are 910 J/kgK,
470 J/kgK and 4200 J/kgK respectively.
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Mass of aluminium = 0.5kg,
Mass of water = 0.2kg
Mass of Iron = 0.2kg
Temp. of aluminium and water = 20°C = 297°k
Sp heat o f Iron = 100°C = 373°k.
Sp heat of aluminium = 910J/kg-k
Sp heat of Iron = 470J/kg-k
Heat again = 0.5 × 910(T – 293) + 0.2 × 4200 × (343 –T)
= (T – 292) (0.5 × 910 + 0.2 × 4200) Heat lost = 0.2 × 470 × (373 – T)
∴ Heat gain = Heat lost
=> (T – 292) (0.5 × 910 + 0.2 × 4200) = 0.2 × 470 × (373 – T)
=> (T – 293) (455 + 8400) = 49(373 – T)
=> (T – 293)(1295/94) = (373 – T)
=> (T – 293) × 14 = 373 – T
=> T = 4475/15 = 298 k
=> T = 298 – 273 = 25°C.
The final temp = 25°C..
hope it's helpful for u
Mass of water = 0.2kg
Mass of Iron = 0.2kg
Temp. of aluminium and water = 20°C = 297°k
Sp heat o f Iron = 100°C = 373°k.
Sp heat of aluminium = 910J/kg-k
Sp heat of Iron = 470J/kg-k
Heat again = 0.5 × 910(T – 293) + 0.2 × 4200 × (343 –T)
= (T – 292) (0.5 × 910 + 0.2 × 4200) Heat lost = 0.2 × 470 × (373 – T)
∴ Heat gain = Heat lost
=> (T – 292) (0.5 × 910 + 0.2 × 4200) = 0.2 × 470 × (373 – T)
=> (T – 293) (455 + 8400) = 49(373 – T)
=> (T – 293)(1295/94) = (373 – T)
=> (T – 293) × 14 = 373 – T
=> T = 4475/15 = 298 k
=> T = 298 – 273 = 25°C.
The final temp = 25°C..
hope it's helpful for u
PriyanshuPrasad:
Did you take approximation in the solution??
Answered by
0
Answer:
the answer is approx 25 °C
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