A 237-g piece of molybdenum, initially at 100.0 C, is dropped into 244 g of water at 10.0 C. When the system comes to thermal equilibrium, the temperature is 15.3 C. What is the specific heat capacity of molybdenum?

The specific heat capacity of molybdenum is

0.2696 joule/gram

Explanation:

Step one

Given that

Mass of molybdenum Mm =237g

Temperature of molybdenum

T1=100°c

Mass of water MW =244g

Temperature of water T2= 10°c

Final temperature T3 =15.3°c

Step two :

We know that the specific heat capacity of water Cw 4.186 joule/gram

But Specific heat capacity of molybdenum Cm?

We know that the quantity of heat transfered can be expressed as

Q=Mm*Cm(T1-T3)+MW*Cw(T2-T3)

since the system attained thermal equilibrium then the expression

Is

=Mm*Cm(T1-T3)=MW*Cw(T3-T2)

Substituting our values into the equation we have

237*Cm(100-15.3)=244*4.186(15.310)

20073.9Cm=5413.3352

Cm=0.2696 joule/gram

Specific heat capacity of molybdenum is 0.2696 joule/gram

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures