- #1

- 3

- 0

1. At 10°C, 32 g of water is mixed with 430 g of ice at -5°C. (The heat capacity of water is 4190 J/(kg * °C), that of ice is 2090 J/ (kg *°C), and the heat of fusion of water is 3.34 * 10^5 J/kg)

a) What mass of water freezes ?

b) What is the change in entropy of the system?

What I did on part (a)

ice : -5°C -> 0°C

Q = m_ice C_ice delta T

Q = (0.43 kg) (2090 J/ (kg *°C) ) (0°C-(-5°C))

Q = + 4493.5 J

water: 10°C -> 0°C

Q = m_water C_water delta T

Q = (0.032 kg) (4190 J/ (kg *°C) ) (0°C-(10°C))

Q = - 1340.8 J

Qice + Qwater = 0

but it doesn't equal to zero.

Not sure what to do next.

Qice + Qwater + Qleftover = 0

Qleftover = m_Water Lf

m_Water = Qleftover / Lf

= 3152.7 J / (3.34 * 10 ^5 J/kg)

=0.00 9439 kg which is approximately 9.4 grams

Doesn't worked out to be the answer of 11 grams.

Where did I gone wrong?

Your help is appreciated.

Thanks.

a) What mass of water freezes ?

b) What is the change in entropy of the system?

What I did on part (a)

ice : -5°C -> 0°C

Q = m_ice C_ice delta T

Q = (0.43 kg) (2090 J/ (kg *°C) ) (0°C-(-5°C))

Q = + 4493.5 J

water: 10°C -> 0°C

Q = m_water C_water delta T

Q = (0.032 kg) (4190 J/ (kg *°C) ) (0°C-(10°C))

Q = - 1340.8 J

Qice + Qwater = 0

but it doesn't equal to zero.

Not sure what to do next.

Qice + Qwater + Qleftover = 0

Qleftover = m_Water Lf

m_Water = Qleftover / Lf

= 3152.7 J / (3.34 * 10 ^5 J/kg)

=0.00 9439 kg which is approximately 9.4 grams

Doesn't worked out to be the answer of 11 grams.

Where did I gone wrong?

Your help is appreciated.

Thanks.

Last edited: