Question

A 920 g empty iron kettle is put on a stove. how much heat in joules musst it absorb to raise its temperature from 15 degrees celcius to 93 degrees celcius? (the specific heat of iron is 113 cal/kg degrees celcius)

Answer 1

To raise the temperature of a 920 g iron kettle from 15 °C to 93 °C, it must absorb 32,369.04 J of heat.Therefore, the kettle must absorb 32,369.04 J to raise its temperature from 15 ℃ to 93 ℃.

Step-by-step explanation:

To determine how much heat is needed to raise the temperature of a 920 g iron kettle from 15 ℃ to 93 ℃, we use the formula:

q = mcΔT

where:

• q is the heat absorbed in joules (J),
• m is the mass of the kettle,
• c is the specific heat capacity of the material,
• ΔT is the change in temperature.

The specific heat capacity of iron given is 113 cal/kg℃, we need to convert it to J/g℃:

113 cal/kg℃ = 0.449 J/g℃ (since 1 cal = 4.184 J and 1 kg = 1000 g)

Now, we can calculate the heat absorbed:

q = (920 g) * (0.449 J/g℃) * (93 ℃ - 15 ℃)

q = (920 g) * (0.449 J/g℃) * (78 ℃)

q = 32,369.04 J