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?

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")))#