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0.5kg of water at 10ºC is completely converted to ice at 0ºC by extracting (88000)...

Physics
JAMB 2021

0.5kg of water at 10ºC is completely converted to ice at 0ºC by extracting (88000) of heat from it. If the specific heat capacity of water is 4200jkg\(^{1}\) Cº. Calculate the specific latent heat of fusion of ice.

  • A. 9.0kjkg\(^{-1}\)
  • B. 84.0kjkg\(^{-1}\)
  • C. 134.0kjkg\(^{-1}\)
  • D. 168.0kjkg\(^{-1}\)
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Correct Answer: Option C
Explanation

Heat involved (H) = 88,000J, Mass(M)= 0.5kg,  specific heat capacity of water(C) = 4200JkgºC,

Specific latent heat of fusion(L) = ?, Temperature change(Δθ) =  θ2 -  θ1 =  (10 - 0)° = 10°

H = MCΔθ + ML 

or

H = M(CΔθ + L) -->\(\frac{H}{M}\) = CΔθ + L 

: L = \(\frac{H}{M}\) -  CΔθ = \(\frac{88,000}{0.5}\) - 4200 \(\times\) 10

L = 176,000 - 42000 = 134,000

L = 134,000 or 134kj/kg

 

There is an explanation video available below.


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Explanation Video

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WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
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WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995