Question 826833
How much heat is required to convert 50g of ice at {{{-10^o}}} Celsius to steam at {{{200^o}}} Celsius?
 
This is a problem that a science teacher (chemistry or physics) would give you.
You need to assume that the process occurs at standard atmospheric pressure (1 atm). (The pressure at start and end is really what matters, and we should assume that is is 1 atmosphere).
You need to calculate 5 different amounts of heat for 5 different stages of the process:
1) the heating of ice from r to {{{0^o}}}{{{C}}} ,
2) the melting at {{{0^o}}}{{{C}}} ,
3) the heating of the resulting liquid water from {{{0^o}}}{{{C}}} to {{{100^o}}}{{{C}}} ,
4) the vaporization of the liquid water at {{{100^o}}}{{{C}}} , and
5) the heating of the resulting water vapor from {{{100^o}}}{{{C}}} to {{{200^o}}}{{{C}}} .
You need data from your textbook, or from a handbook of physics (or chemistry, or engineering).
I tried to find data, and calculate accurately, but you may have more accurate data, or be less prone to calculation mistakes, or your data may be in different units. 
 
Water, as a liquid between {{{0^o}}}{{{C}}} and {{{100^o}}}{{{C}}} ,
has a "specific heat" of about {{{1}}}{{{calorie/g/degreeC}}} .
As ice, a bit below {{{0^o}}}{{{C}}} , or
as vapor at {{{100^o}}} to {{{200^o}}}{{{C}}} ,
it has a "specific heat" of about {{{0.5}}}{{{calorie/g/degreeC}}} .
That is the heat it takes to increase the temperature of {{{1g}}} water by {{{1^o}}}{{{C}}}  (according to my handbook).
(Of course, the number is a function of temperature, but it does no vary by too much).
The heats, or "latent heats", of melting (for ice) and vaporization (for liquid water) are about {{{1440}}} and {{{9700}}}{{{calorie/mol}}} respectively (according to my handbook).
 
So the amounts of heat required are
{{{(50g)*0.5}}}{{{cal/g}}}{{{(0-(-10))^o}}}{{{C}}}={{{250cal}}} to heat the ice to {{{0^o}}}{{{C}}} ,
{{{(50g/"18 g / mol")*1440}}}{{{calorie/mol=4000cal}}}= about {{{4cal}}} to melt the ice at {{{0^o}}}{{{C}}} ,
{{{(50g)*1}}}{{{cal/g}}}{{{(100-0)^o}}}{{{C}}}={{{5000cal}}} to heat the liquid water from {{{0^o}}}{{{C}}} to {{{100^o}}}{{{C}}} ,
{{{(50g/"18 g / mol")*9700}}}{{{calorie/mol=27000cal}}} to vaporize the water at {{{100^o}}}{{{C}}} , and
{{{(50g)*0.5}}}{{{cal/g}}}{{{(200-100)^o}}}{{{C}}}={{{2500cal}}} to heat the water vapor from {{{100^o}}}{{{C}}} to {{{200^o}}}{{{C}}} .
According to my calculation, the total energy in calories is
{{{250+4000+5000+27000+2500=38700}}} .

NOTE:
Your data may slightly different, or it may be in different units (Joules rather than calories, or Joule/g rather than calorie/mol, for example).