Question

Find the result of mixing 10g of ice at -10°C. Specific heat capacity of ice =2.1J/g/K, specific latent heat of ice = 336 J/g and heat capacity of water =4.2J/g/K.

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \purple{ \boxed{ \: \: \: \: \: { \underline{ \purple{ \underline{{ \huge { \mathbb{ \purple{{ \: A}}}}}{ \purple{ \tt{nswer↓ \: \: }}}}}}} \: \: \: \: \: }}}$

Given,

• Mass of ice = 10g
• Temperature of ice = -10°C
• mass of water = 10g
• temperature of water = 10°C
• specific heat capacity of water = 4.2 J/g/K
• specific heat capacity of ice = 2.1 J/g/K
• latent heat of ice = 336 J/g

Let final temperature be t°C.

Concept,

Water(10°C) ⟹ Water(t°C) ⟹ Water(0°C) ⟹ Ice(0°C) ⟹ Ice(-10°C)

In this question,

hot body = water

cold body = ice

according to principle of calorimetry,

Heat released = heat gained

⟹ Heat energy released by water = Heat energy absorbed by ice

when water changes from (10°C to t°C),

${ \green{ \sf { Q = m_w×c_w×∆t}}}$

$\sf \: ⟹ 10g × 4.2 J/g/K × (10 - t)°C$

$\sf \: ⟹ 42 (10 - t)°C$

when ice changes from (-10°C to 0°C),

${ \green{ \sf{Q_1 = m_i × c_i × ∆t}}}$

$\sf \: ⟹ 10g × 2.1 \: J/g/K× 10°C$

$\sf \: ⟹ 201 \: J$

when ice changes to water from (0°C to 0°C of water),

${ \green{ \sf{Q_2 = m_w × L_i}}}$

$\sf \: ⟹ 10g × 336J$

$\sf \: ⟹ 3360 \: J$

when water changes from (0°C to t°C),

Q_3 = m_w × c_w × ∆t

⟹ 10g × 4.2 J/g/K × t°C

⟹ 42t J

Now,

${ \pink { \sf{ ⟹ Q = Q_1 + Q_2 + Q_3}}}$

⟹ 42 (10 - t)°C = 42t + 3360 + 210

⟹ 420 - 42t = 42t + 3360 + 210

⟹ 42 - 3360 - 210 = 42t + 42t

⟹ - 3318 = 84t

We can conclude that the value of t will come in negative, which means that if t is negative then after adding ice to the water, it freezes which is surely not possible. so we will see what's happening here in the following:-

____________________________

Max. heat released by water = mc∆t

⟹ 10 × 4.2 × 10

⟹ 420 J

Heat req. by ice to reach 0°C = mc∆t

⟹ 10 × 2.1 × 10

⟹ 210 J

Heat remaining = 420J - 210J

⟹ 210 J

if all ice melts, then

Heat req. = mL

⟹ 10 × 336

⟹ 3360 J

We can see that it is impossible for all ice to melt as the heat energy released by water < heat energy needed to melt the ice.

So,

If all ice will not melt,

Let (x) g ice be melted,

Q = mL

210 = x × 336

x = 210/336

⟹ x = 0.625 g (ans)

Note:-

The value of ∆t (change in temperature) is always positive whether the answer is coming in negative or not.