You gave me all the info you need.
Do it for one KILOGRAM because those are your units. DIVIDE answer by 1000 then.
Heat the ice from -30 to 0
30 * 2090 J
melt the ice
333000 J
heat the resulting water 100 degrees
100 * 4180 Joules
boil the water
2,260,000 Joules
heat the steam
(120-100)(2,010) Joules
Add them up and divide by 1000 to get a gram instead of a kilogram
Find out the amount of heat required to convert 1 gm of ice at -30 ⁰C to 1 gm of steam at 120
⁰C. Given the specific heat of ice is 2090 J/kg/⁰C, specific heat of water 4180 J/kg/⁰C, specific heat
of steam is 2010 J/kg/⁰C, latent heat of ice to water is 333000 J/kg and latent heat of water to steam
is 2260000 J/kg.
Plz solve it....
2 answers
Hlo
2 answers