Question

How much heat is required to convert 74 grams of ice at -4 C into 74 grams of water at 52 c

Answer 1

1) the first step is to convert ice at -4 C to ice at 0 C

now

heat required to convert ice at -4 C to ice at O C is given by

Q1 = mass of ice x specific heat of ice x temp change

Q1 = m x s x dT

given

mass of ice = 74 g

specific heat of ice = 2.01

temp change = 0- (-4) = 4

so

Q1 = 74 x 2.01 x 4

Q1 = 594.96 J

2)

now the second step is too melt the ice at 0 C to water at O C

so

the heat required to melt the ice at 0 C to water at 0 C is given by

Q2 = mass of ice x heat of fusion of water

Q2 = m x dHfus

given

mass of ice = 74 g

heat of fusion of water = 333.55 J/g

so

Q2 = 74 x 333.55

Q2 = 24682.7 J

3)

now the step is to convert water at 0 C to water at 52 C

so

the heat required to convert water at O C to water at 52 C is given by

Q3 = mass x specific heat of water x temp change

Q3 = m x s x dT

given

mass of water = 74 g

specific heat of water = 4.184 J / g C

temperature change = 52-0 = 52

so

Q3 = 74 x 4.184 x 52

Q3 = 16100.032 J

now

total heat required to convert ice at -4 C to water at 52 C is given by

total heat = Q1 + Q2 + Q3

total heat = 594.96 + 24682.7 + 16100.032

total heat = 41377.692 J

total heat required = 41.377 x 1000 J

total heat required = 41.377 kJ

so

41.377 kJ of heat is required to convert 74 grams of ice at -4 C to 74 grams of water at 52 C