If it takes 1 calorie to raise the temperature of water by 1 degree Celcius, will it take 100 calories to bring the same amount of ice to a boil?

Let's assume we already made all those "ice going to a gaseous state is called sublimation, not boiling"-jokes: You supposedly want to look up the term "latent heat".

The specific heat of water is 4.18*10^3 J / kg K.

 

It takes 4.18*10^3 J to raise the temperature of 1 kilogram by 1 degree (Celsium or Kelvin).

 

If you don't specify how much water you heat with the energy, then you're not making a correct statement.

 

So, the correct question would have been:

If it takes 1 calorie (4.18 J, please use SI units!) to raise the temperature of 1 gram of water by 1 degree Celcius, will it take 100 calories to bring the same amount of ice to a boil?

 

Answer: no. You first need to melt the ice and turn it into water. It would take 100 times the energy to raise the temperature of liquid water of 0 deg C to 100 deg C. Melting ice takes additional energy, so it will be more than 100 times the energy.

 

But that's a physics issue, not math.

I probably overcomplicated the question. Here it is again without ice or vapor considerations.

 

If it takes 1 calorie to raise the temperature of 1 gram of water by 1 degree Celcius, will it take 50 calories to its temperature from 30 degress Celcius to 80 degrees Celcius?

Roughly speaking, yes. However -- What definition of the calorie? There are several definitions, ranging 4.182 to 4.190 joules. One big reason for the different definitions is that the heat capacity of water is not constant. It drops slightly from 0^oC to 34^oC and then rises slightly from 35^oC to 100⁰C. The variation is small but measurable.