Question

If 200. g of water at 20°C absorbs 41 840 J of energy, what will its final temperature be? (Specific Heat of water is 4.184 J/g*C)

Answer 1

Answer: The final temperature will be $70^0C$

Explanation:

To calculate the specific heat of substance during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed =41840 J

c = specific heat = $4.184J/g^0C$

m = mass of water  = 200 g

$T_{final}$ = final temperature =?

$T_{initial}$= initial temperature = $20^0C$

Now put all the given values in the above formula, we get:

$41840J=200g\times 4.184J/g^0C\times (T_{final}-20)^0C$

$T_{final}=70^0C$

Thus the final temperature will be $70^0C$

Answer 2

The final temperature will be $70^oC$

Specific heat capacity

It is defined as the heat capacity of a sample of the substance divided by the mass of the sample.

Formula to calculate specific heat capacity:

• $q=m*C*(T_2-T_1)$

where,

q = heat absorbed =$41840 J$

C = specific heat = $4.184 J/g^oC$

m = mass of water = $200 g$

T₁= initial temperature = $20^oC$

To find:

T₂= final temperature =?

Now put all the given values in the above formula, we get:

$q=m*C*(T_2-T_1)\\\\41840J=200g*4.184J/g^oC(T_2-20^oC)\\\\T_2=70^oC$

Thus, the final temperature will be $T_2=70^oC$

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