Question

a 0.400 kg aluminum teakettle contains 2.00 kg of water at 15.0 degrees celcius. How much heat is required to raise the temperature of the water (and kettle) to 100 degrees celcius

Answer 1

To find the heat required to heat both the aluminum teakettle and water to 100 degrees Celsius, the formula for heat energy considering the mass and specific heat capacity of both substances and the temperature change is used, resulting in a total of 741,880 Joules needed.

Step-by-step explanation:

To calculate the heat required to raise the temperature of both an aluminum teakettle and the water inside it to 100 degrees Celsius, we use the formula:

Q = mwater × cwater × ΔT + maluminum × caluminum × ΔT

where:

Q is the total heat energy,

mwater is the mass of the water,

cwater is the specific heat capacity of water (4.184 J/g°C),

maluminum is the mass of the aluminum kettle,

caluminum is the specific heat capacity of aluminum (900 J/kg°C),

ΔT is the change in temperature.

Since we have been given the masses in kilograms, we need to convert the specific heat capacity of water to J/kg°C by multiplying it by 1000.

Let's plug in the numbers:

Q = (2.00 kg × 4184 J/kg°C × (100°C - 15°C)) + (0.400 kg × 900 J/kg°C × (100°C - 15°C)) = (2.00 kg × 4184 J/kg°C × 85°C) + (0.400 kg × 900 J/kg°C × 85°C) = (2.00 kg × 355,640 J) + (0.400 kg × 76,500 J) = 711,280 J + 30,600 J = 741,880 J

Thus, the total heat energy required is 741,880 Joules.

Answer 2

Q= 742.22 KJ

Step-by-step explanation:

Teakettle mass m_{Al}= 0.400 Kg

mass of water m_{water}= 2 Kg

The initial temperature T_i=15°C

Final temperature T_f=100°C

To find the amount of heat require to raise this temperature

we know that

Q= mcΔT

c= specific heat capacity of material

c for Al= 0.9 KJ/Kg K

c for water = 4.186 KJ/Kg K

m= mass of the material

ΔT= increase in the temperature

Therefore

`Q= m_(Al)c_(Al)\Delta T+m_(water)c_(water)\Delta T`

putting the values we get

`Q=0.4*0.9(100-15)+2*4.186(100-15)`

Q= 742.22 KJ