Camp K.
3 Answers By Expert Tutors
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
Christina W. answered • 07/10/19
Your math and science help is here
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
William W. answered • 07/10/19
BS in Mechanical Engineering with 35 years experience
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 Ti be the initial temperature of the sample and Tf be the final temperature of the sample - we'll use °C in this case.
The governing equation is Q = mC(Tf - Ti) 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.
The key to this problem lies in the value of the specific heat of iron.
#c_"iron" = "0.108 cal"# #color(blue)("g"^(-1))color(darkorange)(""^@"C"^(-1))#
This tells you that in order to increase the temperature of one unit of mass of iron, i.e. of #color(blue)("1 g")# of iron, by one unit of temperature, i.e. by #color(darkorange)(1^@"C")#, you need to provide it with #"0.108 cal"#.
Now, you can use the specific heat of iron to figure out the amount of heat needed to increase the temperature of #"35.0 g"# of iron
#35.0 color(red)(cancel(color(black)("g"))) * "0.108 cal"/(color(blue)(1)color(red)(cancel(color(blue)("g"))) * color(darkorange)(1^@"C")) = "3.78 cal"# #color(darkorange)(""^@"C"^(-1))#
This tells you that in order to increase the temperature of #"35.0 g"# of iron by #color(darkorange)(1^@"C")#, you need to provide it with #"3.78 cal"# of heat.
In your case, the temperature of the iron must increase by
#35^@"C" - 25^@"C" = 10^@"C"#
which means that you will need
#10color(red)(cancel(color(black)(""^@"C"))) * "3.78 cal"/(color(darkorange)(1)color(red)(cancel(color(darkorange)(""^@"C")))) = "37.8 cal"#
Now, you should round the answer to one significant figure, the number of sig figs you have for the change in temperature, i.e. for #10^@"C"#, but I'll leave it rounded to two sig figs
#color(darkgreen)(ul(color(black)("heat needed = 38 cal")))#