- #1

JoshMG

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The question is:

An insulated beaker with negligible mass contains liquid water with a mass of 0.225 kg and a temperature of 75.8 C.

How much ice at a temperature of -23.6 C must be dropped into the water so that the final temperature of the system will be 31.0 C?

Take the specific heat of liquid water to be 4190 J/kg*K, the specific heat of ice to be 2100 J/kg*k, and the heat of fusion for water to be 334 KJ/kg.

mice=?

This is what I have so far:

mw= 0.225 kg

Tw= 75.8 C

Ti= -23.6 C

mi= ?

Tf= 31.0 C

Cw= 4190 J/kg(K)

Ci= 2100 J/kg(K)

Lf= 334 KJ/kg = 334000J/kg (I would have to convert this right?)

With the information given I come up with these:

Qice warms= mi ci (0- (-5))

Qice melts= mi Lf

Qmix of waters= mi cw (31-0)

Qwater cools= mw cw (31 - 75.8)

Qwater cools = Qice warms + Qice melts + Q mix of waters

so now we can solve the mass of ice. right? But when I calculate it the answer I is really weird.

An insulated beaker with negligible mass contains liquid water with a mass of 0.225 kg and a temperature of 75.8 C.

How much ice at a temperature of -23.6 C must be dropped into the water so that the final temperature of the system will be 31.0 C?

Take the specific heat of liquid water to be 4190 J/kg*K, the specific heat of ice to be 2100 J/kg*k, and the heat of fusion for water to be 334 KJ/kg.

mice=?

This is what I have so far:

mw= 0.225 kg

Tw= 75.8 C

Ti= -23.6 C

mi= ?

Tf= 31.0 C

Cw= 4190 J/kg(K)

Ci= 2100 J/kg(K)

Lf= 334 KJ/kg = 334000J/kg (I would have to convert this right?)

With the information given I come up with these:

Qice warms= mi ci (0- (-5))

Qice melts= mi Lf

Qmix of waters= mi cw (31-0)

Qwater cools= mw cw (31 - 75.8)

Qwater cools = Qice warms + Qice melts + Q mix of waters

so now we can solve the mass of ice. right? But when I calculate it the answer I is really weird.

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