Question #157121
A well lagged aluminium calorimeter of mass 80g contains 150g of water and 100g of ice at 0^oC. A heating coil rated 1.0kw is put in the calorimeter and the mixture is stirred until its temperature is 33^oC. Calculate how much ice is left after 1 minute. State any assumption you made.
1
Expert's answer
2021-02-22T10:29:54-0500

Mass of aluminum, ma=80g=0.08Kgm_a = 80g = 0.08Kg

Mass of water, mw=150g=0.15Kgm_w = 150 g = 0.15 Kg

Mass of ice, mi=100g=0.1Kgm_i = 100g = 0.1 Kg

Specific heat of water, sw=4182K/Kg/Ks_w = 4182K/Kg/K

Specific heat of aluminum, sa=0.9J/gK=900J/Kg/Ks_a = 0.9 J/g K = 900J/Kg/K

Specific heat of ice, si=2.04KJ/Kg/K=2040J/Kg/Ks_i = 2.04KJ/Kg/K = 2040J/Kg/K

Latent heat of ice, Lf=3.33×105J/KgL_f = 3.33 \times 10^5 J/Kg

Initial temperature, Ti=0CT_i = 0^\circ C

Final temperature, Tf=33CT_f = 33^\circ C


Heat generated by the coil in one minute, Q=1000×60J=60000JQ = 1000 \times 60 J = 60000 J

Let x amount of the ice that melt,

then assume that there is no loss of the heat,

(masa+mwsw+xsi)(TfTi)+miLf=60000(m_a s_a+m_ws_w+xs_i)(T_f-T_i)+m_iL_f =60000

Putting values,

[(0.080×900)+(0.15×4182)+(x×2040)](330)+(0.1×3.33×105)=60000[(0.080 \times 900)+(0.15 \times 4182)+(x \times 2040)](33-0) +(0.1 \times 3.33 \times 10^5) = 60000


Solving it, we get, x=0.05Kg=50gx =0.05 Kg = 50 g


Amount of ice left, M=10050=50gM = 100-50 = 50 g


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Comments

Tayong Carlson
15.01.22, 13:17

This is really helpful

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