The specific heat of ice is 0.492 cal/ (g × °c). how many calories of heat are required to raise 100.0 g of ice from -20.0°c to -0.5°c?

For this case we have that the number of calories is given by the following equation:

C = mCpΔt

Where,

m: mass

Cp: specific heat

Δt : temperature differential

Substituting values we have:

Around 959.4 calories of heat are required to raise 100.0 g of ice from -20.0 ° c to -0.5 ° c

answer:

noting will happen take it easy ..

explanation:

don't be too much serious or worried nothing will happen to you .

drink a glass of water and maybe . just keep on roaming for a while so that the water is absorbed easily and don't forget to drink plenty of water .okk don't worry you will be fine

ground wouldnt be clean and the soap wouldnt be dirty