Find the result of mixing 10 g of ice at -10° C with 10 g of water at 10° C. Specific heat capacity of ice = 2.1 J g^-1 K^-1, specific latent heat of ice = 336 J g^-1, and specific heat capacity of water = 4.2 J g^-1 K^-1.

Given,

mass of ice = 10 g

mass of water = 10 g

Specific heat capacity of ice = 2.1 J g^-1 K^-1

specific latent heat of ice = 336 J g^-1,

specific heat capacity of water = 4.2 J g^-1 K^-1

Let the final temperature be t° C.

Heat energy taken by ice at – 10° C to raise it's temperature to 0° C (Q₁)
= m x c x change in temperature
= 10 x 2.1 x [0 - (-10)]
= 10 x 2.1 x 10
= 210 J

Hence, Q₁ = 210 J

Heat energy taken by ice at 0° C to convert into water at 0° C (Q₂)
= m x L_ice
= 10 × 336
= 3360 J

Hence, Q₂ = 3360 J

Heat energy taken by water at 0° C to raise it's temperature to t° C (Q₃) = m x c x change in temperature
= 10 × 4.2 × (t – 0)
= 10 × 4.2 × t
= 42 t

Hence, Q₃ = 42 t

Heat energy released by water at 10° C to lower it's temperature to t° C (Q₄) = m x c_water x change in temperature
= 10 × 4.2 × (10 – t)
= 42 × (10 – t) = 420 – 42t

Hence, Q₄ = 420 – 42t

If there is no loss of heat,

Heat energy gained = Heat energy lost

$210 + 3360 + 42 t = 420 - 42t \\[0.5em] 3570 - 420 = - 42 t - 42 t \\[0.5em] 3150 = - 84 t \\[0.5em] t = - \dfrac{3150}{84} \\[0.5em] t = - 37.5° C$

This cannot be true because water cannot exist at -37.5° C.

Hence, we can say that the whole of the ice did not melt.

Let amount of ice which melts = m g

Final temperature of the mixture = 0° C

Heat energy gained by ice at -10° C to raise it's temperature to 0° C

= m x c_ice x change in temperature
= 10 × 2.1 x [0 - (-10)]
= 10 × 2.1 x 10
= 210 J

Heat energy gained by m gm of ice at 0° C to change into water at 0° C
= m × L_ice
= m x 336
= 336m J

Heat energy released by 10 g of water at 10° C to lower it's temperature to 0° C
= m x c_water x change in temperature
= 10 × 4.2 × (10 – 0)
= 10 × 4.2 × 10 = 420 J

If there is no loss of heat,

Heat energy gained = Heat energy lost

$210 + 336 m = 420 \\[0.5em] \Rightarrow 336 m = 420 - 210 \\[0.5em] \Rightarrow 336 m = 210 \\[0.5em] \Rightarrow m = \dfrac{210}{336} \\[0.5em] \Rightarrow m = 0.625 g$

Hence, 0.625 g of ice will melt and temperature will remain at 0° C