6 A radiator made out of iron of specific heat

Berikut ini adalah pertanyaan dari edwardtian123 pada mata pelajaran Fisika untuk jenjang Sekolah Menengah Atas

6 A radiator made out of iron of specific heat capacity 450Jkg-¹K-1 has a mass of 45.0 kg and is filled with 23.0kg of water of specific heat capacity 4200J kg ¹K¹. a Determine the energy required to raise the temperature of the radiator-water system by 1K. b If energy is provided to the radiator at the rate of 450 W, calculate how long it will take for the temperature to increase by 20.0°C.solution please?​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawaban:

1330.44 seconds / 22.17 minutes

Penjelasan:

To determine the energy required to raise the temperature of the radiator-water system by 1K, you need to calculate the total heat capacity of the system. The heat capacity of a substance is the amount of energy required to raise its temperature by 1 degree Celsius.

The heat capacity of the radiator can be calculated using the formula:

Heat capacity = mass * specific heat capacity

For the iron radiator, the heat capacity is 45.0 kg * 450 J/kg/K = 20250 J/K

The heat capacity of the water can be calculated in the same way:

Heat capacity = mass * specific heat capacity

= 23.0 kg * 4200 J/kg/K = 9660 J/K

The total heat capacity of the radiator-water system is the sum of the heat capacities of the iron radiator and the water:

Total heat capacity = 20250 J/K + 9660 J/K = 29910 J/K

Therefore, the energy required to raise the temperature of the radiator-water system by 1K is 29910 J.

To calculate how long it will take for the temperature to increase by 20.0°C if energy is provided to the radiator at the rate of 450 W, you can use the formula:

Time = (energy required) / (rate of energy transfer)

The energy required to raise the temperature of the radiator-water system by 20.0°C can be calculated by multiplying the total heat capacity of the system by the temperature increase:

Energy required = 29910 J/K * 20.0 K = 598200 J

Plugging this value into the time equation above, you get:

Time = (598200 J) / (450 W)

= 1330.44 seconds

= 22.17 minutes

So it will take approximately 22.17 minutes for the temperature to increase by 20.0°C if energy is provided to the radiator at the rate of 450 W.

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Last Update: Thu, 16 Mar 23