Question

Calculate the heat absorbed by 200 g ice present at 0 °C to raise its
temperature to water at 100 °C. [Sp. Heat capacity of water = 4.2 jgł°C,
Sp. Latent heat of ice = 336 jg1]​

Answer 1

Given that,

$Mass\: of\: ice,\:m = 200\; g$

$Change\: in\:temperature,\: \Delta T = 100^{\circ} C$

• Ice present at $0^{\circ} C$ raises its temperature to water at $100^{\circ} C$

$Specific\:heat\:capacity\:of\:water,\:S = 4.2 \;Jg^{-1} {{}^{\circ} C}^{-1}$

$Specific\:Latent \:heat\: of \:ice,\: L = 336\; Jg^{-1}$

We need to find,

Heat absorbed by Ice, Q =?

Formulae to be considered,

• Formula for Heat required to change the temperature

        $Q_S = m S \Delta T$

• Formula for Heat required to change the state of substance

        $Q_L = m L$

Where,

$Q_S = Heat\:required\:for\:temperature\: change, \\Q_L = Heat\: required\: for\: changing\: state, \\\: m = mass\:of\:substance,\: \\\Delta T = change\: in\: temperature, \\ L = specific\:Latent\: heat\: of\:substance, \\\:S = specific \:heat\: capacity\: of \:substance$

So,

For the given case,

→ Total heat abosrbed by ice = Heat required to convert ice into water  + Heat required to change the temperature from $0^{\circ} C$ to $100^{\circ} C$

$\to Q = Q_L + Q_S \\ \\\to Q = m L + m S \Delta T \\ \\\to Q = (200\;g) (336\:Jg^{-1}) + \\\qquad \qquad \qquad \qquad (200\;g)(4.2\;Jg^{-1}{{}^{\circ}C}^{-1})(100^{\circ}C) \\ \\\to Q = 67200\:J + 84000\;J\\ \\\to Q =151200 \:J = 151.200 \;kJ$

Therefore,

Heat absorbed by ice will be 151.200 Kilo-Joules.

Answer 2

$\Large{\bf{\green{\underline{GiVeN,}}}}$

• 200 g ice at 0°C is converted into water at 100°C.

• Specific Heat Capacity of water $\rm{(C_w)}$ is 4.2 J/g.°C.

• Latent heat of ice $\rm{(L_f)}$ is 336 J/g.

$\\ \Large{\bf{\pink{\underline{DiAgRaM,}}}}$

$\bf{\boxed{\orange{0°C\:{\atop{ice}}}}}\:\underset{\red{Q_1}}{\xrightarrow{200\:g}}\:{\boxed{\purple{0°C\:{\atop{water}}}}}\:\underset{\red{Q_2}}{\xrightarrow{200\:g}}\:{\boxed{\green{100°C\:{\atop{water}}}}}$

$\\ \Large{\bf{\blue{\underline{To\:FiNd,}}}}$

• The heat absorbed by ice to converted into water at 100°C.

$\\ \Large{\bf{\purple{\underline{CaLcUlAtIoN,}}}}$

ƇƛƧЄ - 1 ;-

☆ Here we calculate, the heat absorbed by 200g of ice at 0°C to converted into water at 0°C.

➛ Let Q₁ heat is absorbed by 200g of ice at 0°C to converted into water at 0°C.

$\bf\red{We\:know\:that,}$

➣ When phase of a body changes, change of phase takes place at constant temperature [Melting point or Boiling point] and heat released or absorbed is

$\pink\bigstar\:\:{\underline{\green{\boxed{\bf{\purple{Heat\:(Q)\:=\:m\:L}}}}}}$

$\bf\orange{Where,}$

• Heat (Q) = Q₁

• m is the mass of the body = 200g

• L is the latent heat = $\rm{(L_f)}$ = 336 J/g.

$:\implies\:\:\bf{Q_1\:=\:200\times{336}}$

$:\implies\:\:\bf\blue{Q_1\:=\:67200\:J}$

ƇƛƧЄ - 2 ;-

☆ Here we calculate the heat required to convert water at 0°C into water at 100°C.

➛ Let Q₂ heat is required to convert water at 0°C into water at 100°C.

$\bf\pink{We\:know\:that,}$

➣ When a substance does not undergo a change of phase (i.e. liquid remains liquid or solid remains solid), then the amount of heat required to raise the temperature of mass 'm' of the substance by an amount ΔT is

$\red\bigstar\:\:{\underline{\purple{\boxed{\bf{\green{Heat\:(Q)\:=\:m\:C\:\triangle{T}\:}}}}}}$

$\bf\blue{Where,}$

• Heat (Q) = Q₂

• m is the mass of substance = 200g

• C is the specific heat capacity of substance = $\rm{(C_w)}$ = 4.2 J/g.°C.

• ΔT is raise in temperature = 100°C - 0°C = 100°C

$:\implies\:\:\bf{Q_2\:=\:200\times{4.2}\times{100}\:}$

$:\implies\:\:\bf\orange{Q_2\:=\:84000\:J}$

$\Large\bf\red{Now,}$

➛ Q is the total heat required to absorbed by 200g of ice at 0°C to converted into water at 100°C.

$\bf\pink{Thus,}$

$\longmapsto\:\:\bf{Q\:=\:Q_1\:+\:Q_2\:}$

$\longmapsto\:\:\bf{Q\:=\:67200\:+\:84000\:}$

$\longmapsto\:\:\bf{Q\:=\:151200\:J}$

$\longmapsto\:\:\bf\green{Q\:=\:151.2\:KJ}$

━─━─━─━─━─━─━─━─━─━─━─━─━─━─━

$\Large\bf\blue{Therefore,}$

The heat absorbed by 200g of ice at 0°C to converted into water at 100°C is 151.2 KJ.