The mass of the water is 2.41 kilograms.
Given:
The water in which 43,120 cal of energy is placed and its temperature changes to 105 degrees celsius from 87.1 degrees celsius.
To find:
The mass of water:
Solution:
The amount of heat added to water = Q = 43,120 cal
[tex]1 cal = 4.184 J\\\\Q=43,120 cal=43,120\times 4.184 J=180,414.08 J[/tex]
The mass of water = m
The specific heat capacity of the water = c = 4.184 J/g°C
The initial temperature of the water = [tex]T_1=87.1^oC[/tex]
The final temperature of the water =[tex]T_2=105^oC[/tex]
The heat energy added to water to raise its temperature from 87.1°C to 105°C is given by:
[tex]Q=m\times c\times (T-2-T_1)\\\\180,414.08 J=m\times 4.184 J/g^oC\times (105^oC-87.1^oC)\\\\m=\frac{180,414.08 J}{4.184 J/g^oC\times (105^oC-87.1^oC)}\\\\m=2,408.94 g\\\\1 g = 0.001 kg\\\\m=2,408.94 g=2,408.94\times 0.001kg\\\\=2.40894 kg \approx 2.41 kg[/tex]
The mass of the water is 2.41 kilograms.
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