A. mass of 3.00 kg of water at 50 degree C has 1.00 kg of ice at 0 degree C adde

Question

A. mass of 3.00 kg of water at 50 degree C has 1.00 kg of ice at 0 degree C added to it and stirred until thermal equilibrium is reached. Assume all the ice melts. Important: show your work clearly for maximum partial credit! Data: c_ice = 2090 J/kg C degree; c_max = 4186 J/kg C degree; L_fusion = 3.35 times 10^5 J/kg. L_vapor = 2.26 times 10^6 J/kg. Write the single calorimetry equation for this situation in the form Q_gain = Q_loss. (NO numbers apart from temperature are allowed here). Solve this equation for the final equilibrium temperature. (NO numbers apart from temperature are allowed here) Plug numbers into your equation above to find the final equilibrium temperature of the mixture.

Explanation / Answer

mass of water, m = 3 kg
mass of ice, mi = 1 kg
let equilirium temp be T

a) Q gain = mCw(50 - T)
Q loss = mi*Lf + mi*Cw(T - 50)
Q gain = Q loss
mCw(50 - T) = mi*Lf + Mi*Cw(T - 50)
b) T = (mCw*50 - mi*Lf + Mi*Cw*50)/(mCw + MiCw)
c) T = (3*4186*50 - 1*335000 + 1*4186*50)/(3*4186 + 1*4186) = 29.99 deg

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