Question

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?

Answer 1

`"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.