Question

1. How do you do this specific heat equation?

Q = mcΔT

M= 0.1 g

C= 1 calorie/gram∙°C

ΔT = (Final Temp- Initial Temp)

ΔT= (139-132)

ΔT= 7

I'm not understand how the "C" part would be caluclated and incorporated into the equation.

Also, in this lab, I had to use a make-shift calorimeter using two styrafoam cups, 2 nails, and boilding water. These are the results I obtained:

Trial	Temperature of Water (C)
1	132
2	136
3	139

Please help with the following questions!

2. How would doubling the number of iron nails used in this lab affect the overall lab results? Explain.

3. A student heats equal masses of iron (c = 0.45 J/g-C) and aluminum (c = 0.90 J/g-C) to the same temperature and then places the metals into separate beakers along with 50 ml of room-temperature water. Which beaker will have the highest final temperature? Explain.

Answer 1

Following is the Solution and Answer to the first question (i.e. Question - 1 ) of the given Question Set.

Solution:

Since we know, from the given, Trials, the following:

• Initial Temperature (T₁) = 132 ^oC
• Final Temperature (T₂) = 139 ^oC

Therefore........... we get.... T = 7 ^oC

Step - 1:

We know the following:

• Specific Heat (i.e. C or c ) = 1 Calorie/ gram. ^oC
• Mass (i.e. M) = 0.1 g (i.e. grams)
• Change/ Increase in Temperature: i.e.  T = 7 ^oC
• Change in Heat (i.e. Q ) = ?

​Therefore...............

Step - 2: ......and .... ( Answer )

Therefore.......

Q = M x C x  T = (0.1 g ) x ( 1 Calorie/ gram. ^oC) x ( 7 ^oC ) = 0.7 Calories ..............

Therefore.....Answer:   Q = 0.7 Calories

Note*:   "C" part is the Specific Heat, defined as the amount of heat (Q) required to raise/change the Temperature of the given substance by 1 ^oC ... which has been provided in the question, and you don't need to calculate.... Hope this would help...