Question

An insulated beaker with negligible mass contains a mass of 0.350 kg of water at a temperature of 69.4 ∘C.

How many kilograms of ice at a temperature of − 14.7 ∘C must be dropped in the water to make the final temperature of the system 29.3 ∘C?

Take the specific heat for water to be 4190 J/(kg⋅K) , the specific heat for ice to be 2100 J/(kg⋅K) , and the heat of fusion for water to be 334 kJ/kg .

Answer #1

When ice at -14.7^{o}C is dropped into water and brought to
a temperature of 29.3^{o}C, it undergoes 3 state changes,
i.e. releases heat in 3 forms. -

1) Temperature of ice rising from
-14.7^{o}C to 0^{o}C = Q = **m _{ice}
x c_{ice} x (T_{f}-T_{i}) =**
m

= 30,870 J/kg x m_{ice}

2) Latent heat of fusion released by
ice = Q_{f} = m_{ice} x L_{f} =
m_{ice} (334000)

3) Temperature of water rising from
0^{o}C to 29.3^{o}C = Q = m_{ice}**x
c _{water} x (T_{f}-T_{i}) =**
m

= 122,767 J/Kg x m_{ice}

Therefore, total heat released by ice should be equal to that absorbed by water,

Q_{ice =}
Q_{water}

=> **m _{ice} x
c_{ice} x (T_{f}-T_{i}) +**
m

**=>** 30,870 J/kg x
m_{ice} + m_{ice} (334000) + 122,767 J/Kg x
m_{ice} = (0.350) x (4190) x(69.4-29.3)

=> **m _{ice} =
(58,806.65)/ (487637) = 0.1206 kg**

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