A. 4.80 kJ
B. 20.16 kJ
C. 40.32 kJ
D. 22.17 kJ
Correct Answer:
Option C = 40.32 kJ
Explanation
Heat needed to change 60g (0.06kg) of ice at 0oC to water at 0o
= 0.06 x 3.36 x 105J
= 0.2016 x 105 J
Heat required to raise the temp. of 60g (0.06kg) of water from 0oC to 80oC
= 0.06 x 4.2 x 103 x (80.0) J
= 0.06 x 4.2 x 103 x 80
= 20.16 x 103
= 0.2016 x 105
Total heat needed = (0.2016 + 0.2016) x 105J
= 0.4032 x 105 J
= 40.32KJ
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![Calculate the energy required to vapourize 50g of water initially at 80°C.[specific heat capacity of water = 2260Jg‾¹K‾¹] [specific latent heat of vaporization of water = 2260Jg‾¹]](https://erudites.ng/wp-content/plugins/contextual-related-posts/default.png "Calculate the energy required to vapourize 50g of water initially at 80°C.[specific heat capacity of water = 2260Jg‾¹K‾¹] [specific latent heat of vaporization of water = 2260Jg‾¹]")