Question

2. A 25.0-g sample of an alloy is heated to 100.0 C and dropped into 90.0 g water at 25.32 C. The temperature of the water rises to 27.18 C. What is the specific heat of the alloy?

Answer 1

@abb0t

Answer 2

q1=-q2
now what?

Answer 3

Solve for \(c\)!

Answer 4

how?

Answer 5

You already have all the information for \(Q_2\), but you're missing some of the information for \(Q_1\), which is \(c\) and you know \(Q = mc \Delta T\), so...plug in the given information and solve for c!

Answer 6

\(Q_{water} = m_{water}c_{water}\Delta T_{water}\)
\(Q_{alloy} = m_{alloy}c_{alloy}\Delta T_{alloy}\)

\(Q_{water} = -Q_{alloy}\)

Answer 7

wait but how do I find c?

Answer 8

That means: \(mc\Delta T_a = -mc \Delta T_b\), NOW, USE the properties that you learned in algebra to rearraange the formula.

Answer 9

that means, \(divide\)...

Answer 10

Ta=Tb

Answer 11

a=-b

Answer 12

\(c_{water} \) is a known value.
You know both masses, and you know both changes in temprature. The only unknown is \(c_{alloy}\).

Answer 13

\(\large c = -\frac{mc \Delta T}{m\Delta T}\)

Answer 14

0.092 cal/g oC @abb0t

Answer 15

sound's about right.

Answer 16

idk if ur untis are correct tho..