Question

How many calories are required to raise the temperature of a 35.0 g sample of iron from 25°C to 35°C? Iron has a specific heat of 0.108 cal/g°C?

Answer 1

`"38 cal"`

Explanation:

The key to this problem lies in the value of the specific heat of iron.

`c_"iron" = "0.108 cal"` `color(blue)("g"^(-1))color(darkorange)(""^@"C"^(-1))`

This tells you that in order to increase the temperature of one unit of mass of iron, i.e. of `color(blue)("1 g")` of iron, by one unit of temperature, i.e. by `color(darkorange)(1^@"C")`, you need to provide it with `"0.108 cal"`.

Now, you can use the specific heat of iron to figure out the amount of heat needed to increase the temperature of `"35.0 g"` of iron

`35.0 color(red)(cancel(color(black)("g"))) * "0.108 cal"/(color(blue)(1)color(red)(cancel(color(blue)("g"))) * color(darkorange)(1^@"C")) = "3.78 cal"` `color(darkorange)(""^@"C"^(-1))`

This tells you that in order to increase the temperature of `"35.0 g"` of iron by `color(darkorange)(1^@"C")`, you need to provide it with `"3.78 cal"` of heat.

In your case, the temperature of the iron must increase by

`35^@"C" - 25^@"C" = 10^@"C"`

which means that you will need

`10color(red)(cancel(color(black)(""^@"C"))) * "3.78 cal"/(color(darkorange)(1)color(red)(cancel(color(darkorange)(""^@"C")))) = "37.8 cal"`

Now, you should round the answer to one significant figure, the number of sig figs you have for the change in temperature, i.e. for `10^@"C"`, but I'll leave it rounded to two sig figs

`color(darkgreen)(ul(color(black)("heat needed = 38 cal")))`