What will be the final temparature of mixture if 1 g of ICE AT 0 DEGREE CELSIUS AND 1 G OF STEAM AT 100 DEGREE CELSIUS.

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Solution

When two different phases are mixed together we calculate by,

m₁C_P1(T_h - T_f) =m₂C_P2(T_f - T_L)
where,
Here C_P1 is specific heat capacity of Steam = 1.996 J/K-g
C_P2 is the specific heat capacity of Ice = 2.108 J/K-g

T_h = 100^oC = 373 K
T_f is the final temperature after mixing
T_L= 0 ^oC = 273 K

m₁C_P1(T_h - T_f) =m₂C_P2(T_f - T_L)
$1 \times 1.996 \times (373 - T_{F}) = 1 \times 2.108 (T_{F} - 273) 1319.98 = 4.104 T_{F} T_{F} = 321.63 K T h e r e f o r e t h e f i n a l t e m p e r a t u r e o f t h e m i x t u r e i s 321.63 K$

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