2019-11-21T13:27:05-05:00
If mr.M has a cup of water with room temperature of 20 °C but adds 80 g of ice to the drink, so it cools to 6°C and adds more ice to drink of water so the temperature is 10°C what will the volume of the drink be?
1
2019-11-22T11:14:51-0500
1.
c M ( T w − T 1 ) = c m ( T 1 − T i ) + m L cM(T_w-T_1)=cm(T_1-T_i)+mL c M ( T w − T 1 ) = c m ( T 1 − T i ) + m L
M ( 4200 ) ( 20 − 6 ) = 80 ( 4200 ( 6 − 0 ) + 334000 ) M(4200)(20-6)=80(4200(6-0)+334000) M ( 4200 ) ( 20 − 6 ) = 80 ( 4200 ( 6 − 0 ) + 334000 ) Initial mass of water:
M = 489 g M=489\ g M = 489 g 2.
c ( M + m ) ( T 2 − T 1 ) = c m ′ ( T w − T 2 ) c(M+m)(T_2-T_1)=cm'(T_w-T_2) c ( M + m ) ( T 2 − T 1 ) = c m ′ ( T w − T 2 )
( M + m ) ( T 2 − T 1 ) = m ′ ( T w − T 2 ) (M+m)(T_2-T_1)=m'(T_w-T_2) ( M + m ) ( T 2 − T 1 ) = m ′ ( T w − T 2 )
( 489 + 80 ) ( 10 − 6 ) = m ′ ( 20 − 10 ) (489+80)(10-6)=m'(20-10) ( 489 + 80 ) ( 10 − 6 ) = m ′ ( 20 − 10 )
m ′ = 228 g m'=228\ g m ′ = 228 g Total mass:
M + m + m ′ = 489 + 80 + 228 = 797 g M+m+m'=489+80+228=797\ g M + m + m ′ = 489 + 80 + 228 = 797 g Total volume is
v = 0.797 ≈ 0.8 L v=0.797\approx0.8\ L v = 0.797 ≈ 0.8 L
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