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 Lice
= 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 cwater 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
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 cice 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 × Lice
= 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 cwater 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
Hence, 0.625 g of ice will melt and temperature will remain at 0° C