Question

How much heat is required to change a 47.2 g ice cube from ice at -11.9°C to steam at 112°C? (if necessary, use c_ice=2090 J/kg°C and c_steam= 2010 J/kg°C)

Answer 1

Useful information:
heat of fusion of water = 334 J/g
heat of vaporization of water = 2257 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.18 J/g·°C
specific heat of steam = 2.09 J/g·°C

Step 1: Heat required to raise the temperature of ice from -11.9 °C to 0 °C Use the formula

q1 = mcΔT = 47.2*2.09*( 0--11.9) = 1173.9112 J

Step 2: Heat required to convert 0 °C ice to 0 °C water

q2= m·ΔH_f = 47.2 *334 = 15764.8 J

Step 3: Heat required to raise the temperature of 0 °C water to 100 °C water

q3 = mcΔT =  47.2*4.18* ( 100-0) = 19729.6 J

Step 4: Heat required to convert 100 °C water to 100 °C steam

q4 = m·ΔH_v = 47.2* 2257 = 106530.4 J

Step 5: Heat required to convert 100 °C steam to 112 °C steam

q5 = mcΔT = 47.2* 2.09* ( 112-100) =1183.776 J

Heat_Total = Heat_{Step 1} + Heat_{Step 2} + Heat_{Step 3} + Heat_{Step 4} + Heat_{Step 5}  

= 1173.9112+ 15764.8+19729.6+106530.4+1183.776 = 144382.487 J answer

_{Good LUck !!}