Question

How much heat is required to convert 3 kg of ice at -12 *C to steam at 100*C specific heat of ice is 2100 j/kg*C and specific heat of water is 4186J/kg*C . Latent heat of fusion is 3.35 * 10^5 J/Kg and latent heat of steam is 2.25*10^6 J/kg*C

Answer 1

Answer:

Total heat (Q) = 9.1 × $10^{6}$ J

Explanation:

We have to convert ice at -12°C ---> steam at 100°C

This conversion will take place in the following manner :-

ɪ ice at -12°C -----> ice at 0°C

ɪɪ ice at 0°C -----> water at 0°C

ɪɪɪ water at 0°C -----> water at 100°C

ɪV water at 100°C -----> steam at 100°C

Let the heat supplied be Q1 , Q2, Q3 and Q4 respectively in the process ɪ, ɪɪ, ɪɪɪ and ɪV.

So,

NOTE :-

The formula for heat when only the state changes but the temp. remains constant is :-

m L                

[m = mass of the water/ice , L= Latent heat of fusion/steam]

and,

The formula for heat when the state remains the same but the temperature varries is :-

mcΔT

[m= mass of water/ice , c = specific heat of water/ice , ΔT = change in temperature]

Now,

Q1 = mcΔT

    = 3 × 2100 × [0 - (-12)]

    = 3 × 2100 × 12

    = 75600 J

    = 0.756 × $10^{5}$ J

Q2 = mL

     = 3 × 3.35 × $10^{5}$

     = 10.05 × $10^{5}$ J

Q3  = mcΔT

      = 3 × 4186 × 100

      = 12558 × 100

      = 12.5 × $10^{5}$ J

Q4 = mL

     = 3 × 2.25 × $10^{6}$

     = 6.75 × $10^{6}$ J

     = 67.5 × $10^{5}$ J

Now,

Q = Q1 + Q2 + Q3 + Q4

   = 75600 + 1005000 + 1250000 + 6760000

   = 9090600

   = 9.1 × $10^{6}$ J

Thanks !

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