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The key to this problem lies in the value of the 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 you have for the change in temperature, i.e. for $10^@"C"$, but I'll leave it rounded to two sig figs

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

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