Question

A 237.0 g sample of molybdenum metal is heated to 100.10 °C and then dropped into an insulated cup containing 244.0 g of water at 10.00 °C. If the final temperature of the water and metal in the cup is 15.30 °C, then what is the specific heat of molybdenum? (Specific heat of water = 4.186 J/g-°C Do not add the unit in the answer.

Answer 1

Answer : The specific heat of molybdenum metal is,
`0.269J/mole^oC`

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`c_1` = specific heat of molybdenum metal = ?

`c_2` = specific heat of water =
`4.186J/g^oC`

`m_1` = mass of molybdenum metal = 237.0 g

`m_2` = mass of water = 244.0 g

`T_f` = final temperature of water and metal =
`15.30^oC`

`T_1` = initial temperature of molybdenum metal =
`100.10^oC`

`T_2` = initial temperature of water =
`10.00^oC`

Now put all the given values in the above formula, we get

`237.0g* c_1* (15.30-100.10)^oC=-244.0g* 4.186J/g^oC* (15.30-10.00)^oC`

`c_1=0.269J/g^oC`

Therefore, the specific heat of molybdenum metal is,
`0.269J/mole^oC`