2 kg Soluti Calculate the mass of liquid water. m water = 30.0 kg Write the equa

Question

2 kg
Soluti Calculate the mass of liquid water.
mwater = 30.0 kg Write the equation of thermal equilibrium. Qice + Qmelt +Qice - water + Qwater =0(1) Construct a comprehensive table. Q m (kg) c (J/kg·°C) L (J/kg) Tf (°C) Ti (°C) Expression Qice 2.00 2,090 0 -4.80 micecice(Tf- Ti) Qmelt 2.00 3.33 105 0 0 miceLf Qice - water 2.00 4,190 T 0 micecwat(Tf- Ti) Qwater 30.0 4,190 T 20.0 mwatcwat(Tf- Ti) Substitute all quantities in the second through sixth columnsinto the last column and sum (which is the evaluation of Equation1), and solve for T. 2.01 104 J + 6.66105 J + (8.38103 J/°C)(T - 0°C) +(1.26 105)(T - 20.0°C) = 0
T = 1°C
Your answer differs from the correct answerby 10% to 100%.
Remarks Making a table is optional. However,simple substitution errors are extremely common, and the tablemakes errors much less likely.

Explanation / Answer

Process : Here ice will gain heat energy and water will loseenergy mass of water = density * volume = 1000*1020 = 1.02 *106 Kg Heat loss by water = mSw(T) where, m = mass of water = 1.02 * 106 Kg = 1.02 *109 g Sw = 4.186 joule/gram K T = change in temperature = 20 - 10 = 10o C =10K => heat loss by water = 1.02 * 109 *4.186 * 10 = 42.697 * 109 J Heat gain by Ice = mSi(T) + mL + mSw(T) where, m = mass of ice = ? Sw = specific heat of ice = 2.06 J / gKL = latent heat of ice = 333.55 J/g => heat gained by ice = m*2.06*(0-(-10.8)) + m*333.55 +m*4.186*(10-0) By priciple of calorimetry Heat loss = heat gain just solve this equation and you will get your answer

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