I place a 250-g ice cube (initially at 0°C) in a Styrofoam box with wall thickness 1.0 cm and total surface area 600 cm2. If the air surrounding the box is at 20°C and after 2 hours the ice is completely melted, what is the conductivity of the Styrofoam material? (Lf = 80 cal/g)Select one:a.1.15 × 10−2 b.9.6 × 10−5 c.2.8 × 10−6 d.2.3 × 10−4
Question
I place a 250-g ice cube (initially at 0°C) in a Styrofoam box with wall thickness 1.0 cm and total surface area 600 cm2. If the air surrounding the box is at 20°C and after 2 hours the ice is completely melted, what is the conductivity of the Styrofoam material? (Lf = 80 cal/g)Select one:a.1.15 × 10−2 b.9.6 × 10−5 c.2.8 × 10−6 d.2.3 × 10−4
Solution
To solve this problem, we need to use the formula for heat conduction, which is Q = kA(T1-T2)*t/d, where:
Q is the heat transferred, k is the thermal conductivity, A is the surface area, T1 and T2 are the initial and final temperatures, t is the time, and d is the thickness of the material.
First, we need to calculate the amount of heat needed to melt the ice. This can be done using the formula Q = m*Lf, where m is the mass of the ice and Lf is the latent heat of fusion. Substituting the given values, we get Q = 250g * 80 cal/g = 20000 cal.
Next, we substitute the known values into the heat conduction formula and solve for k:
20000 cal = k * 600 cm^2 * (20°C - 0°C) * 2 hours / 1.0 cm
Solving for k, we get k = 20000 cal / (600 cm^2 * 20°C * 2 hours) = 8.33 * 10^-5 cal/cm°Cs.
However, we need to convert the time from hours to seconds because the standard unit for time in the thermal conductivity formula is seconds. There are 3600 seconds in an hour, so 2 hours is 7200 seconds.
So, k = 8.33 * 10^-5 cal/cm°Cs * 1 hour/3600 s = 2.31 * 10^-6 cal/cm°Cs.
Therefore, the thermal conductivity of the Styrofoam is approximately 2.31 * 10^-6 cal/cm°Cs, which is closest to option c. 2.8 × 10−6.
Similar Questions
A box with a total surface area of 1.20 m2 and a wall thickness of 4.00 cm is made of an insulatingmaterial. A 10.0-W electric heater inside the box maintains the inside temperature at 15.0°C abovethe outside temperature. Find the thermal conductivity k of the insulating materi
1. What is the thermal conductivity of: and provide several factors (applicable one or several of these products) that could impact the thermal conductivity values you provide. a. Dow Styrofoam SM® extruded polystyrene insulation b. EnergyGuard Polyiso c. EnergyGuard™ NH Polyiso d. Roxul CavityRock® stonewool insulation e. R-22 EcoTouch® PINK® FIBERGLAS® Insulation f. Grizzly Gold Medium Density closed cell spray foam.
A thermos flask with evacuated space to reduce the heat losses havingsurfaces facing each other of emissivity 0.02. If contents of the flaskare at 380 K & the ambient temperature is 298 K. Compute the heatloss from the flask. If same effect is to be achieved by using insulatingmaterial of conductivity 0.04 w/m K. What would be the thicknessrequired
Which of the following solutions will have the highest conductivity at 298 K?(a) 0.01 M HCl solution (b) 0.1 M HCl solution(c) 0.01 M CH 3COOH solution (d) 0.1 M CH 3COOH solution
A cylinder is made up of two coaxial layers, one of radius R and the other of inner radius R and outer radius 2R. The inner and the outer portions are respectively made up of substances of thermal conductivities K and 2K respectively. If the effective thermal conductivity between the flat ends of the cylinder is xk/4 , then the value of x is ________
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.