Question

An insulated beaker with negligible mass contains a mass of 0.200 kg of water at a temperature of 81.7 ∘C.

How many kilograms of ice at a temperature of − 21.1 ∘C must be dropped in the water to make the final temperature of the system 30.8 ∘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

The water cools from 81.7° to 30.8° Celsius. That is a change of
50.9°C. This is the same scale as Kelvin so the change is 50.9
K

Multiply (0.20 kg)x(50.9 K)x(4190 J/(kg⋅K)) = 42654.2 Joules lost
by the water.

The final temperature of the ice is 30.8°C,

To find certain mass (M) = Mx(30.8 K)(4190 J/(kg⋅K)) is the
amount of energy from freezing to 37°C. ------(1)

and Mx(3.34 x 10^{5} J/kg) is the energy during
melting.-----(2)

Then Mx(21.1 K)(2100 J/(kg⋅K)) is the energy from -21.1°C to 0°C. --------(3)

Add equations (1)+(2)+(3)

=>(Mx30.8x4190)+(Mx3.34x10^{5})+(Mx21.1x2100)=42654.2

129052M+334000M+44310M=42654.2

Mx507362=42654.2

M=0.008kg of ice

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