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## Homework Statement

An ice cube at

**0.00 °C**measures

**14.9 cm**on a side. It sits on top of a copper block with a square cross section

**14.9 cm**on a side and a height of

**18.1 cm**. The bottom of the copper block is in thermal contact with a large pool of water at

**92.5 °C**. How long does it take the ice cube to melt? Assume that only the part in contact with the copper liquefies, that is, the cube gets shorter as it melts. The density of ice is

**0.917 g/cm**.

^{3}T

_{ice}= 273 K

T

_{water}= 365.5 K

L

_{Heat of fusion}= 334 kJ/kg

V

_{ice}= (14.9cm)

^{3}= 3,307.95 cm

^{3}

h = 0.181 m

κ

_{Cu}= Thermal conductivity of Copper = 385 W/(m*K)

A = face area of Cu = (0.149m)

^{2}= 0.0222 m

^{2}

## Homework Equations

ρ = m/V

Q = mL

_{Heat of fusion}

P

_{Cond}= Q/

*t*= Aκ(ΔT/L)

## The Attempt at a Solution

I did this is 3 steps.

1)

Used density to obtain the volume of the ice cube

ρ

_{i}= m

_{i}/V

_{i}

m

_{i}= ρ

_{i}V

_{i}

m

_{i}= (0.917 g/cm

^{3})(3,307.95 cm

^{3})

m

_{i}= 3,033.39 g

2)

Used the mass to obtain amount of energy required to melt the ice, Q

_{i}.

Q

_{i}= m

_{i}L

_{Heat of fusion}

Q

_{i}= (3,033.39g)(334 J/g)

Q

_{i}= 1,013,152.3 J

3)

Used Q

_{i}and the conduction rate equation to calculate the time

*t*.

P

_{Cond}= Q

_{i}/

*t*= A

_{Cu}κ

_{Cu}(ΔT/L

_{Cu})

Q

_{i}L

_{Cu}= (A

_{Cu}κ

_{Cu}ΔT)

*t*

Q

_{i}L

_{Cu}/(A

_{Cu}κ

_{Cu}ΔT) =

*t*

*t*= (1,013,152.3 J)(0.181 m)/(0.0222 m

^{2})(385 W/(m*K))(92.5 K)

*t*= 231.952 J/W = 231.952 (W*s)/W = 231.952 s

*t*[itex]\approx[/itex] 232 s

I submitted this answer to my online homework and it told me I was incorrect. I need to make sure that I am correctly analyzing the situation and accounting for everything I need to.

Any help is appreciated.

Please don't be rude. I will gladly provide more info on my calculations. Thanks.