usy02al.theexpertita com Class Management I Help hermo start ofoscillations Begi
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usy02al.theexpertita com Class Management I Help hermo start ofoscillations Begin Date: 412/2017 12:00.00 AM Due Date: 419/2017 500:00 PM End Date: 5/2/2017 12200000 AM (8%)Problem 4: Several ice cubes (ou se 09167 gem of total volume vi 225 cm' and temperature 2m3.15 K (0.000 ac) are put into a thermos containing ve 680 cm3 of tea at a temperature of 313.15 K completely filling the thermos. The lid is then put on the thermos to close it.Assume that the density and the specific heat of the tea is the same as itis for fresh water (e. 100gom 33% Part (a) Calculate the amount of heat energy emin Jneeded to melt the ice cubes (L1 klkg. Grade Summary sino coso tano 7 8 9 cotano asino acos0 atano acotano sinho oDegrees Radians Feed ack deduction per ferdheck Hints: deduction per hint. Hints nomaining 33 Parta) calculate the equilibrium temperature T in K of the final mixture of tea and water.Explanation / Answer
mass of Ice, m_Ice = density*volume
= 0.9167*200
= 183 grams
= 0.183 kg
a) amount of heat required to melt the ice,
Qm = m_Ice*Lf
= 0.183*334*10^3
= 61122 J
b) mass of tea, m_Tea = 0.68 kg
let T is the final equilibrium temperature.
heat gained by Ice = Heat lost by Tea
m_Ice*Lf + m_Ice*C_water*(T - 273.15) = m_Ice*C_water*(313.15 - T)
0.183*334*10^3 + 0.183*4186*(T -273.15) = 0.68*4186*(313.15 - T)
==> T = 287.7 K
c) Heat lost by tea = m_Ice*C_water*(313.15 - T)
= 0.68*4186*(313.15 - 287.7)
= 72443 J
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