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?

2 Answers
Jul 7, 2018

Answer:

Well, your ice DOES NOT undergo a phase change....

Explanation:

And so we simply take the product...

#"mass"xx"C"_"specific heat"xxDeltaT=100.0*gxxunderbrace(0.492*cal*g^-1*""^@C^-1)_"quoted specific heat"xx19.5*""^@C=+959.4*cal#....the positive sign indicates heat is ADDED to the system.

Jul 7, 2018

Answer:

#959.4# calories

Explanation:

We use the specific heat equation, which states that,

#q=mcDeltaT#

where:

  • #q# is the heat energy supplied in joules

  • #m# is the mass of the substance in kilograms

  • #c# is the specific heat capacity of the object in joules

  • #DeltaT# is the change in temperature

Here, #DeltaT=-0.5^@"C"-(-20^@"C")=19.5^@"C"#.

So, we get:

#q=100color(red)cancelcolor(black)"g"*(0.492 \ "cal")/(color(red)cancelcolor(black)"g"color(red)cancelcolor(black)(""^@"C"))*19.5^@"C"#

#=959.4 \ "cal"#