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  • I need help with A metal ball of mass 0.1 kg is heated upto 500 oC and dropped into a vessel of heat capacity 800 JK-1 and containing 0.5 kg water . The initial temperature of water and vessel is 30 oC. What is the approximate percentage increment in the

A metal ball of mass 0.1 kg is heated upto 500 oC and dropped into a vessel of heat capacity 800 JK-1 and containing 0.5 kg water . The initial temperature of water and vessel is 30 oC. What is the approximate percentage increment in the temperature of the water ? [ Specific Heat Capacities of water and metal are, respectively, 4200 Jkg-1K-1 and 400 4200 Jkg-1K-1 ]

  • Option 1)

     25%

  • Option 2)

    15 %

  • Option 3)

    30 %

  • Option 4)

    20 %

Answers (1)

best_answer

 

Specific Heat -

C=\frac{Q}{m.\Delta \theta}

- wherein

C = specific heat

\Delta \theta=Change in temperature 

m = Amount of mass

 

s_{m}=400 9kg^{-1}k^{-1}

s_{w}=4200 9kg^{-1}k^{-1}

apply heat lost = heat gain

0.1\times 400\times (500-T)=0.5\times 4200\times (T-30)+800(T-30)

\Rightarrow 40(500-T)=(T-30)[800+2100]

20\times 1000-40T=2900T-(2900\times 30)

\Rightarrow T=36.4

So, \frac{\Delta T}{T}\times 100=\frac{6.4}{30}\times 100

\approx 20%%

 

 


Option 1)

 25%

Option 2)

15 %

Option 3)

30 %

Option 4)

20 %
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