If a wire of resistance 20 ohm is covered with ice at zero degree Celsius and voltage of 210 V is applied across the wire then the rate of melting of ice is
If a wire of resistance 20 ohm is covered with ice at zero degree Celsius and voltage of 210 V is applied across the wire then the rate of melting of ice is
#If a wire of resistance of 20 ohms is covered with ice at zero degree Celsius and voltage of 210 V is applied across the wire then the rate of melting of ice is?
Ans: Resistance R of the Wire= 20 Ohm
Voltage drop V across the resistance = 210 Volt
Wattage W of the resistance wire = V²/R = (210×210)/20 = 2205 Joule/sec
Watt = 2205 Joule/sec= (2205 Joule/second)÷ (4.1858 Joule/ calorie)
= 526. 78 calorie/sec
The rate of heat produced in the resistance wire = 526.78 calories per second.
Each gram of water takes 80 calories of heat to melt and change to water at 0°C.
The rate of melting of ice =(526.78 calories per second)÷(80 calories per gram)
= 6.58 g of ice per second.
Ice would be melting at the rate of 6.58g per second due to the current flowing in the resistance wire.
Ans:
Resistance R of the Wire= 20 Ohm
Voltage drop V across the resistance = 210 Volt
Wattage W of the resistance wire = V²/R = (210×210)/20 = 2205 Joule/sec
Watt = 2205 Joule/sec= (2205 Joule/second)÷ (4.1858 Joule/ calorie)
= 526. 78 calorie/sec
The rate of heat produced in the resistance wire = 526.78 calories per second.
Each gram of water takes 80 calories of heat to melt and change to water at 0°C.
The rate of melting of ice =(526.78 calories per second)÷(80 calories per gram)
= 6.58 g of ice per second.
Ice would be melting at the rate of 6.58g per second due to the current flowing in the resistance wire.
