Question

how much heat os required to raise ths temperature of 100gm of water from 5°c to 95°​

Answer 1

Answer:

9 calories or 37.674 joules

Explanation:

Heat required

= heat absorbed by water while its temperature raises from 5°C to 95°C

= mass of water * specific heat of water * change in temperature

= 100 g * 4.186 J/kg°C * (95°C - 5°C)

= 0.1 kg * 4.186 J/kg°C * 90°C

= 37.674 J (or) 9 cal

Answer 2

Answer:

specific heat capacity of water= 4.2 J/g C

quantity of heat = 4.2 *100*(95-5)

                             =4.2*100*90

                              =37800 J

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