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storchak [24]
3 years ago
13

if 980 KJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be

Chemistry
1 answer:
uranmaximum [27]3 years ago
6 0

Answer:

The final temperature of the water will  be 328.81 K .

Explanation:

Using the equation, <em>q = mcΔT</em>

here, <em>q = energy</em>

<em>m= mass</em>

<em>c= specific heat capacity </em>

<em>ΔT= change in temperature</em>

<em>Mass of water = 1kg (1000 g ) per liter</em>

<em>∴ 6.2 Liter of water = 6200 g</em>

<em>c of water ≈ 4.18 J /g/K</em>

<em>Now, </em>

<em>980000 = 6200*4.18*ΔT</em>

<em>ΔT = 37.81 K</em>

<em>∴ final temperature of the water = 291 + 37.81 = 328.81 K</em>

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On a warm summer day, a large mass of air (atmospheric pressure 1:01 105 Pa) is heated by the ground to 32.0 C and then begins t
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The process is adiabatic because there no heat transfer into the system or out of the system in that there is only one system involved in this process that is air.

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   T_{final}  =  278.5 \ K  

Explanation:

From the question we are told that

      The atmospheric pressure is P_a  = 1.01 *10^{5} \ Pa

       The temperature heated to T  = 32.0^oC =  305 \ K  

       The new atmospheric pressure is  P_a__{n}}  =  8.0 * 10^{4} \ Pa

       The value of g is  g =  1.40

        The rate of cooling is  r =  \frac{1}{100}  \ ^oC/m

The adiabatic relation for this process is mathematically represented as

            P^{1- g}_{initial} T^{g}_{initial} =  P^{1-g} _{final} T^{g}_{final}

Now   P}_{initial}  =  P_ a ,   T_{initial} =  T,P _{final} =  P_a__{n}} ,  T_{final}  = ?

substituting value  

           (1.01*10^{5})^{1-1.40} * (305) ^{1.40} = (8.0*10^{4})^{1-1.40} * T_{final} ^{1.40}

          T_{final} ^{1.40}  =  2646.7

 =>    T_{final} ^{\frac{1.40}{1.40} }  =  2646.7^{(\frac{1}{1.40})}        

        T_{final}  =  278.5 \ K  

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3 years ago
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