How many calories are required to raise the temper ature of a 35.0 g sample of iron from 25 °C to 35 °C? Iron has a specific heat of 0.108 cal/g °C

Answer 1

Since you have two answers that differ, I thought I'd weigh in and "break the tie", so to speak.

q = mC∆T

q = heat = ?

m = mass = 35.0 g

C = specific heat = 0.108 cal/g/degree

∆T = change in temperature = 35-25 = 10 degrees

q = (35.0 g)(0.108 cal/g/deg)(10 deg)

q = 37.8 cal

PS. Just noticed that William W used 108 cal/g/deg instead of 0.108 cal/g/deg

Answer 2

How many calories are required to raise the temper ature of a 35.0 g sample of iron from 25 °C to 35 °C? Iron has a specific heat of 0.108 cal/g °C

Camp hope you are doing well. The equation to solve this is Q=mcΔt

Q = energy (Joules or Calories)

m= mass

C = specific heat

ΔT = change in temperature

Q = (35.0g)(0.108c/g℃)(35℃ - 25℃)     1. I always find my change first per PEMDAS from

Math.

Q= (35.0g)(0.108c/g℃)(10℃) 2. Now keep in mind that c/g℃ is a fraction and we

can view as cg℃ Now when we do this

we can see that the grams can cancel and the can cancel

each other out so you are left with just calories.

Q=(35.0)(0.108c)(10) 3. Now just multiply through.

Q = 37.8 cal

Answer 3

Let Q be the amount of heat required (the units of heat can be calories or joules or other units of energy - we'll pick calories in this case)

Let m be the mass of the sample in question (the units of mass can be grams or kilograms or other units of mass - we'll pick grams)

Let C be the Specific Heat of the material the sample is made of - this is a physical property of the material you look up in a book - in this case, you have the number and it is 0.108 cal/g °C.

Let T_i be the initial temperature of the sample and T_f be the final temperature of the sample - we'll use °C in this case.

The governing equation is Q = mC(T_f - T_i) so, plugging in the numbers:

Q = (35g)(0.108 cal/g °C)(35°C - 25°C) = 37.8 calories.

Since the problem inputs have two sig figs for the temperatures, we would round the answer to two sig figs or 38 calories.