Question

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?

Answer 1

The specific heat capacity of molybdenum is

0.2696 joule/gram

Step-by-step 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