Ice of mass 11.0 kg at 0.00° C is placed in an ice chest. The ice chest has 3.00 cm thick walls of thermal conductivity 1.00 10-5 kcal/s · m · C° and a surface area of 1.25 m2.
(a) How much heat must be absorbed by the ice before it melts?
(b) If the outer surface of the ice chest is at 29.0° C, how long will it take for the ice to melt?
Q = r•m =335000•11 = 3.685•10^6 J,
where r is the heat of fusion.
q = α•(ΔT/Δx)•A,
where q is the heat flow per unit time,
α is åðó thermal conductivity,
α =1•10^5 kcal/s • m • C° =
=0.041868 J/s • m • C°,
ΔT is the temperature driving force,
Δx is the thickness of the walls,
and A is the surface area,
q = 0.041868•(29/0.03) •1.25 =
=50.59 J/s.
t = Q/q = 3.685•10^6/50.59 = 7.28•10^4 s = 20.23 h.
To answer these questions, we need to calculate the following:
(a) The amount of heat absorbed by the ice before it melts.
(b) The time it takes for the ice to melt.
Let's start by solving part (a) first.
(a) How much heat must be absorbed by the ice before it melts?
To determine the amount of heat absorbed by the ice, we need to use the formula:
Q = mcΔT
where Q is the amount of heat absorbed or released, m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature.
Given:
m = 11.0 kg (mass of the ice)
ΔT = 0.00°C (change in temperature)
To find the value of c (specific heat capacity of ice), we can refer to a table of physical constants. The specific heat capacity of ice is approximately 2.09 J/g°C (joules per gram per degree Celsius).
However, the mass is given in kilograms, so we need to convert it to grams.
m = 11.0 kg × 1000 g/kg = 11000 g
Now we can calculate the amount of heat absorbed by the ice:
Q = mcΔT
= 11000 g × 2.09 J/g°C × 0.00°C
= 0 J
Therefore, the amount of heat absorbed by the ice before it melts is 0 joules.
Now let's move on to part (b).
(b) If the outer surface of the ice chest is at 29.0°C, how long will it take for the ice to melt?
To calculate the time it takes for the ice to melt, we need to consider the heat flow through the walls of the ice chest.
The rate of heat flow (Q/t) can be calculated using Fourier's law of heat conduction:
Q = kAΔT/t
where Q is the amount of heat flow, k is the thermal conductivity of the material (given as 1.00 × 10^(-5) kcal/s · m · °C), A is the surface area of the ice chest (given as 1.25 m^2), ΔT is the temperature difference between the inner and outer surfaces of the ice chest, and t is the time it takes for the ice to melt (the value we want to find).
Given:
k = 1.00 × 10^(-5) kcal/s · m · °C
A = 1.25 m^2
ΔT = 0.00°C - 29.0°C = -29.0°C
We can rearrange the formula to solve for t:
t = kAΔT / Q
= (1.00 × 10^(-5) kcal/s · m · °C) × (1.25 m^2) × (-29.0°C) / 0 J
However, we encounter a division by zero, indicating that the ice will not melt under the given conditions. The amount of heat flow required to melt the ice is 0 J, as calculated in part (a). Moreover, the negative temperature difference indicates that heat is flowing from the colder ice to the warmer surroundings, which is not possible.
Therefore, the ice will not melt in the scenario described, and the time it takes for the ice to melt is undefined.