An aluminum calorimetry cup [mc = 0.6 kg, cc = 0.22 kcal/(kgC)] is filled with water [mw = 2.4 kg, cw = 1 kcal/(kgC)]. It is at a temperature of 30C. A sample of an unknown substance with mass mm = 2.34 kg, originally at a temperature of 210C, is added. After a while, the temperature of the water is 35C. What is the specific heat of this material?

Question

An aluminum calorimetry cup [mc = 0.6 kg, cc = 0.22 kcal/(kgC)] is filled with water [mw = 2.4 kg, cw = 1 kcal/(kgC)]. It is at a temperature of 30C. A sample of an unknown substance with mass mm = 2.34 kg, originally at a temperature of 210C, is added. After a while, the temperature of the water is 35C. What is the specific heat of this material?

Answer 1

I guess both the cup and the water started at 30 C and ended at 35 C

so they went up 5 C

heat in = .6*.22*5+2.4*1*5 = 12.66 KCal

heat out = 2.34 * c * (210-30)
= 421 c KCal

heat in = heat out

12.66 = 421 c

c = 0.03

are you sure you added 2.34 kg of this stuff and not maybe 2.34 grams?

Answer 2

To find the specific heat of the unknown substance, we can use the principle of energy conservation:

The heat lost by the unknown substance when it cools down will be equal to the heat gained by the water and the aluminum cup when they warm up.

Let's go step by step:

1. Calculate the heat gained by the aluminum cup and the water:
The heat gained by the aluminum cup is given by:
Q1 = mc * cc * (Tf - Ti)
where
mc = mass of the aluminum cup = 0.6 kg
cc = specific heat of aluminum = 0.22 kcal/(kgC)
Tf = final temperature = 35C
Ti = initial temperature = 30C

Q1 = 0.6 kg * 0.22 kcal/(kgC) * (35C - 30C)
Q1 = 0.6 * 0.22 * 5 kcal

The heat gained by the water is given by:
Q2 = mw * cw * (Tf - Ti)
where
mw = mass of water = 2.4 kg
cw = specific heat of water = 1 kcal/(kgC)
Tf = final temperature = 35C
Ti = initial temperature = 30C

Q2 = 2.4 kg * 1 kcal/(kgC) * (35C - 30C)
Q2 = 2.4 * 1 * 5 kcal

2. Calculate the heat lost by the unknown substance:
The heat lost by the unknown substance is given by:
Q3 = mm * cm * (Tf - Ti)
where
mm = mass of the unknown substance = 2.34 kg
Tf = final temperature = 35C
Ti = initial temperature = 210C (converted from Celsius to Kelvin: 210C + 273 = 483K)

Now, we can equate the heat gained by the aluminum cup and the water to the heat lost by the unknown substance:
Q1 + Q2 = Q3

0.6 * 0.22 * 5 kcal + 2.4 * 1 * 5 kcal = 2.34 * cm * (35C - 210C)

Simplifying the equation:
0.66 kcal + 12 kcal = -2.34 * cm * 175C
12.66 kcal = -409.5 * cm

Solving for cm:
cm = -(12.66 kcal) / (409.5 * 175C)

Evaluating this expression will give you the specific heat of the material.