Silver and Copper rods of equal areas are placed end to end with the free end of the silver rod in ice at 0.00 degrees Celsius and the free end of the copper rod in steam at 100. degrees Celsius. The Silver rod is 15.0 cm in length and the copper rod is 25.0 cm in length.

a) What is the temperature of the junction between copper and sliver when they have come to equilibrium?

b) How much ice (in grams) melts per second?

A.) The temperature of the junction between copper and sliver when they have come to equilibrium is 35 degree Celsius

B.) ice (in grams) melts per second = 0.078 kg/s

Explanation:

A.) Given that the two material are of the same area.

The Silver rod is 15.0 cm in length and the copper rod is 25.0 cm in length.

Silver temperature = 0 degree Celsius

Copper temperature = 100 degree Celsius

Thermal conductivity k of silver = 429 W/m•K

Thermal conductivity k of copper = 385W/m. k

Rate of energy transferred P in the two materials can be expressed as

P = k. A. dT/L

dT = change in temperature

Since the rate and the area are the same

429 (T - 0) / 0.15 = 385 (100 - T) / 0.25

2860T = 1540 (100 - T)

Open the bracket

2860T = 154000 - 1540T

Collect the like terms

2860T + 1540T = 154000

4400T = 154000

T = 154000/4400

T = 35 degree Celsius

The temperature of the junction between copper and sliver when they have come to equilibrium is 35 degree Celsius

B.) Rate of energy transferred P will be

P = 2860 * 35 = 100100

P = Q/t ... (1)

Where Q = energy transferred

But Q = mcØ ... (2)

And specific heat capacity c of water = 4182J/k. kg

Substitutes Q into formula 1.

P = mcØ/t

Make m/t the subject of formula

m/t = P/cØ

m/t = 100100 / 4182 (35 + 273)

m/t = 100100/1288056

m/t = 0.078 kg/s