How Does Changing Shape Affect Temperature and Ice Thickness Calculation?

[Zekexx]

Homework Statement

Question 1: A solid sphere has a temperature of 873 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature in Kelvin?

Questions 2: A 0.454-m-thick sheet of ice covers a lake. The air temperature at the ice surface is -13.4 °C. In 1.03 minutes, the ice thickens by a small amount. Assume that no heat flows from the ground below into the water and that the added ice is very thin compared to 0.454 m. Find the number of millimeters by which the ice thickens.

Homework Equations

Question 1 : We are to assume that the radius of the sphere is R and side length of the cube is L and that volume is kept constant so L = (4/3pi)^1/3R and Q/t cube = Q/t sphere the areas of and cue is 6L^2 and a sphere is 4piR^2

Question 2 : First we must find the amount of heat flow from Q=((KAT)t)/L now the A for this problem is 1 m^2 after Q is found we can use Q=ML to find mass and density = MV to find the volume.

The Attempt at a Solution

Question 1 : Since everytihng is read to be kept constant except for temperature and area can't we just condense so that T^4A = T^4A and than we can substitute in for L so that we get
873^4*4*pi*R^2 = T^4*6*(4/3pi)^2/3*R^2
this would allow our R's to cancel and then it is just algebra...is this right or am i missing something?

Question 2:
Q=((KAT)t)/L
Q=(2.2*1*13.4*(1.03*60))/.454

Q = ML
So Q/L=M

and Density = M*V

so Density/M = V

am i missing something in this part of the problem?

Thanks to any who answer

 

[Redbelly98]

Everything looks right except your equation for density. Think carefully about that one again.

p.s. Welcome to Physics Forums