How are coffee cups used to calculate the specific heat of a metal

Specific heat capacity of metal is 2.09 j/g.°C.

Explanation:Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of metal = 15.5 g

Initial temperature = 165.0°C

Initial temperature of water = 23.0°C

Final temperature = 30.0°C

Specific heat capacity of metal = ?

Specific heat capacity of water = 4.184 J/g°C

Volume of water = 150.0 mL or 150.0 g

Solution:

Formula:

- Qm  =  +Qw

Now we will put the values in formula.

-15.5 g × c × [ 30.0°C - 165.0°C] = 150 g × 4.184 J/g°C × [ 30.0°C - 23.0°C]

15.5 g × c × 135°C = 4393.2 j

2092.5 g.°C  × c = 4393.2 j

c = 4393.2 j/2092.5 g.°C  

c = 2.09 j/g.°C  

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'