# 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?

Aug 12, 2016

$\text{10 cal}$

#### 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 $\text{1 g}$ of said substance by ${1}^{\circ} \text{C}$.

For ice, you know that if you provide $0.5$ calories of heat to $\text{1 g}$ of ice you will increase its temperature by ${1}^{\circ} \text{C}$.

This is how much heat you need to supply to a given sample of ice for every gram and for every ${1}^{\circ} \text{C}$ increase in temperature.

To increase the temperature of $\text{20 g}$ of ice by ${1}^{\circ} \text{C}$, you need to provide it with $\text{0.5 cal}$ for every gram. This will cause its temperature to increase by ${1}^{\circ} \text{C}$.

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

For a ${1}^{\circ} \text{C}$ 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 ${1}^{\circ} \text{C}$ you need to provide it with $\text{10 cal}$ of heat for every ${1}^{\circ} \text{C}$ increase in temperature.

For example, to increase the temperature of $\text{20 g}$ of ice by ${3}^{\circ} \text{C}$, 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 $\text{20 g}$ of ice by ${3}^{\circ} \text{C}$, you need to provide it with $\text{30 cal}$ of heat.