How many kilocal are required to increase the temperature of 15.6 g of iron from 122 °C to 355 °C? The specific heat of iron is 0.450 J/g °C
no the real answer is 0.391 kcal
1.64 ^ is wrong.
Let me help you with that. The correct answer is 3.91 kcal.
To determine the amount of energy required to increase the temperature of a substance, we can use the formula:
q = mcΔT
where:
q = energy (in joules)
m = mass (in grams)
c = specific heat capacity (in J/g °C)
ΔT = change in temperature (in °C)
First, we need to convert the mass of iron from grams to kilograms. Since there are 1000 grams in 1 kilogram, the mass of the iron is:
m = 15.6 g / 1000 = 0.0156 kg
Next, we can calculate the change in temperature:
ΔT = 355 °C - 122 °C = 233 °C
Now we can plug in the values into the formula:
q = (0.0156 kg) x (0.450 J/g °C) x (233 °C)
Simplifying the equation:
q = 0.0156 kg x 0.450 J/g °C x 233 °C
Since the units of grams and °C cancel each other out:
q = 0.0156 kg x 0.450 J/g x 233
Calculating the value:
q = 0.0156 kg x 0.450 J/g x 233 = 1.62936 J
Since the specific heat capacity is given in joules per gram, we need to convert the answer to kilojoules:
q = 1.62936 J / 1000 = 0.00163 kJ
Therefore, it requires approximately 0.00163 kilocalories (kcal) to increase the temperature of 15.6 g of iron from 122 °C to 355 °C.