Question

 What quantity of heat is needed to melt 120 g of ice at -20 o

c to water at 30 o

c? Specific 

heat capacity of water and ice are 4200 JKg-1K

-1 and 2100 JKg-1K

-1

respectively. And 

also to specific latent heat 0f ice as 3.36 x 105

JK-1

Answer 1

We can find the quantity of heat from the formula:

$$Q = Q_1+Q_2+Q_3,$$

here, $Q_1 = mc_{ice} \Delta T_1$ is the quantity of heat needed to change the temperature of the ice from $-20 \ ^{\circ}C$ to $0 \ ^{\circ}C$, $Q_2 = mL_f$ is the quantity of heat needed to transform ice into the water at $0 \ ^{\circ}C$, $Q_3 = mc_{water} \Delta T_3$ is the quantity of heat needed to change the temperature of the water from $0 \ ^{\circ}C$ to $30 \ ^{\circ}C$, $m = 0.12 \ kg$ is the mass of the substance, c_{ice} = 2100 \ \dfrac{J}{kg \cdot \! ^{\circ}C} is the specific heat capacity of ice, c_{water} = 4200 \ \dfrac{J}{kg \cdot \! ^{\circ}C} is the specific heat capacity of water, $L_f = 3.36 \cdot 10^5 \ \dfrac{J}{kg}$ is the latent heat of fusion of ice.

Then, we get:

$$Q = mc_{ice} \Delta T_1+mL_f+mc_{water} \Delta T_3,$$

Q = 0.12 \ kg \cdot(2100 \ \dfrac{J}{kg \cdot \! ^{\circ}C} \cdot 20 \ ^{\circ}C +3.36 \cdot 10^5 \ \dfrac{J}{kg}+4200 \ \dfrac{J}{kg \cdot \! ^{\circ}C} \cdot 30 \ ^{\circ}C),

$$Q = 60480 \ J.$$

Answer:

$Q = 60480 \ J$.