Question

Calculate the final temperature of a mixture of 110.00 g of ice initially at -7.00 °C and 286.50 g of water initially at 88.50 °C

Answer 1

Solution- Here, qgain = -qlost

Ice is melting from -7.0 C to 0 C:
Now using formula-
q = cmdT

here,
c = specific heat of ice = 2.108 J/gC
m = mass = 110.0 g
dT = change in temperature = (0 - (-7))

q = 2.108 * 110 * 7 = 1623.16 J


Ice melting:

q = Hf * m
Hf = heat of fusion = 334 J/g for ice
m = 110 g

q = 334 * 110 = 36740 J


The melted ice heating up from 0 C to final temperature, x:

q = cmdT
c = specific heat = 4.184 J/gC for water
m = mass = 140 g
dT = change in temperature = (x - 0)

q = 4.184 * 110 * x = 460.24x

qgain = qtotal = 1623.16 + 36740 + 460.24x = 38363.16 + 460.24x



The hot water is losing heat to final temperature, x:

q = cmdT

q = 4.184 * 286.50 * (x - 88.5)


qgain = -qlost

38363.16 + 460.24x  = -(4.184 * 286.50 * (x - 88.5))

38363.16 + 460.24x= -1198.72x+106086.37

1658.96x=67723.21

= 67723.21 /1658.96

=40.82 C

x = 38.6 C