Question

You have 1.30 kg of water at 27.6 Degree Celsius in an insulated container of negligible mass. You add 0.580 kg of ice that is initially at -21.0 Degree Celsius. Assume no heat is lost to the surroundings and the mixture eventually reaches thermal equilibrium. If all of the ice has melted, What is the final temperature (in Degree Celsius, round to 2 decimal places) of the water in the container? Otherwise if some ice remains, what is the mass of ice (in kg, round to 3 decimal places) that remains?

NB. L_{f,ice} = 3.34*10^{5} J/kg,
c_{water} = 4190 J/kg*K, c_{ice} =
2100J/kg*K

Answer #1

Solution :

Here we have given :

m_{water} = 1.30 kg

Initial temperature of water : T_{w} = 27.6
^{o}C

m_{ice} = 0.580 kg

Initial temperature of ice : T_{i} = - 21
^{o}C

.

Here, Heat lost by the water as its temperature changes from
27.6 ^{o}C to 0 ^{o}C will be :

Q = m_{water} C_{water} ΔT = (1.30 kg)(4190 J/kg
^{o}C )(27.6 ^{o}C - 0 ^{o}C) = 150337.2
J

.

Let the amount of ice that were melted is
**m**.

Then : m_{ice} C_{ice} (0 - T_{i}) +
**m** L_{f} = Q

∴ (0.58 kg)(2100 J/kg K)(21 ^{o}C) + **m**
(3.34 x 10^{5} J/kg) = 150337.2 J

∴ **m** (3.34 x 10^{5} J/kg) = 124759.2
J

∴ **m** = 0.374 kg

.

Therefore, The mass of ice that remains will be =
m_{ice} - **m** = 0.580 kg - 0.374 kg =
**0.206 kg**

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