A 50-g ice cube at 0oC is heated until 45-g hasbecome water at 100oC and 5.0-g has become steam at100oC. How much energy was added to accomplish thetransformation?

Melt the ice cube, turning it into 0 degwater 
Heat the water from 0 to 100 
Turn 5 grams of the water into steam. 
#1 melting ice cube,phase change Q = m L or Q = 50 grams * 79.7 cal/gram = 3985 calories 
#2, heating the water: Q = m c?T or Q = 50 grams* 1 cal/gram C * 100 C = 5000calories 
#3, turning 5 grams into steam... another phasechange: Q = mL Q = 5 grams * 540 cal/gram = 2700 calories 

3985 + 5000 + 2700 = 11,685 calories 
In Joules, 11,685 cal* (4.19 Joule/ 1 cal ) = 48,960 joules

First, it has to be made clear that the exercise must have a typing error because it states that the 50 kilogram ice cube is heated until 45 g of it turns into water at boiling temperature and 5 g into steam at the boiling point. This would imply that either all of the cube's mass entered the liquid phase at boiling point and only 5 g of it evaporated, or that we somehow divided 50 g and ignored the other. We'll presume that there is an error and take the mass of the ice cube to be 50 g based on the total of the figures. Let's go on.

A melting process of 50 g of ice, the heat of which is $Q_m$ , a process of heating up of the entire 50 g of water until boiling point, the heat of which is $Q_h$ , and the process of evaporating of 5 g of water at boiling point, the heat of which is $Q_v$. Let's denote the total (50 g) mass by $m_v$. 

$Q_m=m{L}_f$

$Q_h=cm\mathrm{△}T$

$Q_v=m_v{L}_v$

$Q=Q_m+Q_h+Q_v$

$Q=m{L}_f+cm\mathrm{△}T+m_v{L}_v=m(L_f+c\mathrm{△}T)+m_v{L}_v$

$Q=0.05\cdot (3.3\cdot {10}^5+4186\cdot 100)+0.005\cdot 2.26\cdot {10}^6=4.87\cdot {10}^{4}J$