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?

The problem provides you with the 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.