A tool smith is making a metal object. The object is to be put into boiling wate

Question

A tool smith is making a metal object. The object is to be put into boiling water in order to cool off the water to 70 degree C. The object must be able to soak up a lot of heat without being too heavy. The tool smith has two metals to work with, A and B. First she puts a 15 g piece of metal A into boiling water and measures how much heat this piece absorbs in rising from 20 degree C to 70 degree C. Then she uses larger and larger piece of A, each time dropping the piece into different amounts of boiling water and finding out how much heat the piece absorbs in rising from 20 degree C to 70 degree C. Eventually, she runs out of metal A and starts using pieces that are combination of metal A and metal B. The graph below shows the heat absorbs by the pieces in rising from 20 degree C to 70 degree C plotted against the mass of the pieces. . A. What are the specific heats of metals A and B? B. How much of each substance was added in all? C. What is the heat capacity of the final combination?

Explanation / Answer

Heat Q = m.S. where m is mass, S is specific heat and change in temperature.

The graph shows heat absorbed Vs mass of metal, first few points are taken when only metal A is used afterwards both metal A and B are used.

We took the first two point to find specific heat of metal A

For point (15,50)

50 = 15 x 50 SA since = 70 -20 = 50

SA = 0.067 cal.g-1K-1   cgs unit

For point (40,150)

150 = 40 x 50 SA

SA = 0.075 cal.g-1K-1

The specific heat of metal A is average of previous two values = 0.0708 cal.g-1K-1

Using this value to find the heat for mass 100g we get H = 100 x 0.0708 x 50 = 354 cal which is almost same as shown in graph.

The other two values are very different from calculated value of heat, which means she had above 100g of metal A but below 180g and then she started adding metal B.

taking the last two point (180,550) (260,650) to find specific heat of metal B

H = mASA + mBSB

The mass of A is fixed in both case since we don't have more metal A.

Therefore,

550 = [mASA + (180-mA)SB ] x 50 and 650 = [mASA + (260-mA)SB] x 50

550 = [mA x 0.0708 + (180-mA) SB ] x 50 and 650 = [mA x 0.0708 + (260-mA) SB] x 50

11 = mA x 0.0708 + (180-mA) SB and 13 = mA x 0.0708 + (260-mA) SB

SB =(11 - mA x 0.0708 ) / (180-mA)

using it we get 13 = mA x 0.0708 + (260-mA) x [(11 - mA x 0.0708 ) / (180-mA)]

By solving the above equation we get,

mA = 141.92 (it might be seen as quadratic but while solving the uadratic term gets cancelled)

and SB = 0.025 cal.g-1K-1

B) finally 141.92 g of metal A and 118.08 g of metal B was added.

C) The heat capacity of final combination was

650 = 260 x Scomb x 50

Scomb = 0.05 cal.g-1K-1

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.