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Question:

An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9C

How much ice at a temperature of −10.0C must be dropped into the water so that the final temperature of the system will be 26.7C?

Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.

**Working out:**

Q_b = m_beaker * c_water * change in temp. of beaker

Q_b = (0.205)(4190)(26.7-79.9) = -45696.14J

Q_1 = m_ice * c_ice * change in temp. of ice

Q_1 = m_ice(2100)(26.7- -10) = m_ice(77070)

Q_2 = m_ice * L_f

Q_2 = m_ice(334000)

Q_b + Q1 + Q2 = -45696.14 + m_ice(77070) + m_ice(334000) = 0

Solving this for m_ice, I get: 0.111kg

I'm told that the answer is wrong with a hint saying "Remember that the ice turns into liquid water above the melting point."

I'm not sure how to go about this, can someone help me out, isn't that hint above the working out I did for Q_2 ?