Question

How much energy is required to change a 35-g ice cube from ice at ?15°C to steam at 115°C?

Answer 1

The first process is change of ice at -15 degrees C to ice at 0 C

So, energy required in this = a = mass in kg*specific heat of ice * temperature change=  .035 * 2090 * 15 = 1097.25J

Similarly, now while conversion of ice at 0C to water at 0C ; energy required = b = mass in kg*latent heat of
fusion of water = 0.035* 333000= 11655J

now while change water at 0 to water at 100 C, c = mass in kg*specific heat of water * temperature change = 0.035*4186*100= 14651 J

now, while change water at 100 C to steam at 100 C, d = mass in kg*latent heat of vaporization = 0.035*2260000= 79100 J

and now, while there is change of steam at 100 C to steam at 115 C , e = mass in kg*specific heat of steam * temperature change = 0.035*2010*15= 1055.25J

So, when added a to e, the energy required to change a 35-g ice cube from ice at ?15°C to steam at 115°C is 1097.25J + 11655J+ 14651 J+ 79100 J+ 1055.25J = 107558.5 J