Answer:
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
