The specific heat of ice is 0.5 calories/gram°C? 20 grams of ice will require how many calories to raise the temperature 1°C?

1 Answer
Aug 12, 2016

Answer:

#"10 cal"#

Explanation:

The problem provides you with the specific heat of ice, which is said to be equal to

#c_"ice" = "0.5 cal g"^(-1)""^@"C"^(-1) = 0.5color(white)(a) "cal"/("g" ""^@"C")#

As you can see, specific heat is expressed in units of energy, in this case calories, per gram Celsius, which means that a substance's specific heat tells you how much energy is needed to increase the temperature of #"1 g"# of said substance by #1^@"C"#.

For ice, you know that if you provide #0.5# calories of heat to #"1 g"# of ice you will increase its temperature by #1^@"C"#.

This is how much heat you need to supply to a given sample of ice for every gram and for every #1^@"C"# increase in temperature.

To increase the temperature of #"20 g"# of ice by #1^@"C"#, you need to provide it with #"0.5 cal"# for every gram. This will cause its temperature to increase by #1^@"C"#.

#20 color(red)(cancel(color(black)("g"))) * ("0.5 cal"""^@"C"^(-1))/(1color(red)(cancel(color(black)("g")))) = "10 cal"""^@"C"^(-1)#

For a #1^@"C"# increase in temperature, you have

#1 color(red)(cancel(color(black)(""^@"C"))) * "10 cal"/(1color(red)(cancel(color(black)(""^@"C")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("10 cal")color(white)(a/a)|)))#

As you can see, to increase the temperature of this sample by more than #1^@"C"# you need to provide it with #"10 cal"# of heat for every #1^@"C"# increase in temperature.

For example, to increase the temperature of #"20 g"# of ice by #3^@"C"#, you need to provide it with

#3 color(red)(cancel(color(black)(""^@"C"))) * overbrace("10 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^color(blue)("amount of heat needed for 1"""^@"C increase") = "30 cal"#

Therefore, you can say that in order to increase the temperature of #"20 g"# of ice by #3^@"C"#, you need to provide it with #"30 cal"# of heat.