A 0.600-kg aluminum teakettle contains 4.00 kg of water at 26.0° C. How much heat is required to raise the temperature of the water (and kettle) to 100.0° C?

Q = c(Al)•m(Al)•Δt + c(wat)•m(wat)• Δt =

=(930•0.6 +4180•4) •74 =1,278,572 J= =1.2•10^6 J = 1.2 MJ

To calculate the amount of heat required to raise the temperature of the water (and kettle) to 100.0° C, we can use the specific heat formula:

q = mcΔT

where:
q = heat energy (in joules),
m = mass of the object,
c = specific heat of the material,
ΔT = change in temperature.

First, let's find the heat required to raise the temperature of the water to 100.0° C:

The specific heat of water is approximately 4186 J/kg°C.

The mass of the water is 4.00 kg.

ΔT = 100.0°C - 26.0°C = 74.0°C.

So, the heat required to raise the temperature of the water is:

q_water = (4.00 kg) × (4186 J/kg°C) × (74.0°C)
= 1239448 J

Next, let's calculate the heat required to raise the temperature of the teakettle:

The specific heat of aluminum is approximately 900 J/kg°C.

The mass of the teakettle is 0.600 kg.

ΔT = 100.0°C - 26.0°C = 74.0°C.

So, the heat required to raise the temperature of the teakettle is:

q_teakettle = (0.600 kg) × (900 J/kg°C) × (74.0°C)
= 39960 J

Finally, to find the total heat required to raise the temperature of both the water and teakettle, we add the two values together:

Total heat required = q_water + q_teakettle
= 1239448 J + 39960 J
= 1279408 J

Therefore, the amount of heat required to raise the temperature of the water (and kettle) to 100.0° C is 1279408 Joules.