Answers
mi = mass of ice = 100 g
M = mass of water = 1000 g
ci = specific heat of ice = 2.108
cw = specific heat of water = 4.186
Tii = initial temperature of ice = - 8
Twi = initial temperature of water = 20
T = final equilibrium temperature
L = latent heat of fusion of ice = 334
using conservation of heat
heat gained by ice = heat lost by water
mi ci (0 - Tii ) + mi L + mi cw (T - 0 ) = M cw (20 - T )
(100 x 2.108) (0 - (- 8)) + (100 x 334) + (100 x 4.186)(T - 0 ) = (1000 x 4.186) (20 - T )
T = 10.6 C
there is only water at temperature of 10.6 C
b)
W = work done
m = mass of water = 100 + 1000 = 1100 g
c = specific heat = 4.186
T = change n temperature = 20 - 10.6 = 9.4 C
work done is same as the heat required to raise the temperature of water
hence
W = Q
W = m c T
W = (1100) (4.186) (9.4)
W = 43283.24 J
C)
m = mass of the weight
W = potential energy of the weight
W = mgh
43283.24 = m (9.8 x 1.4)
m = 3154.76 kg
m = 3.15 tonnes
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