Question

4 OT 5

A 35g chunk of metal at 130C was dropped in a bucket containing 220g of water at 25C. The final temperature of the mixture
was 300. What is the specific heat capacity of the metal?
(Specific heat for water: 4184 J/g C) Formula (m)(c)(T) = (m) (c)(T)

Answer 1

Answer: The specific heat capacity of metal is
`1.31J/g^0C`

Step-by-step explanation:

`Q_(absorbed)=Q_(released)`

As we know that,

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c* (T_(final)-T_1)=-[m_2* c* (T_(final)-T_2)]`

where,

`m_1` = mass of metal = 35 g

`m_2` = mass of water = 220 g

`T_(final)` = final temperature =
`30^0C`

`T_1` = temperature of metal =
`130^oC`

`T_2` = temperature of water =
`25^oC`

`c_1` = specific heat of metal = ?

`c_2` = specific heat of water =
`4.184J/g^0C`

Now put all the given values in equation (1), we get

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]`

`35* c_1* (30-130)^0C=-[220g* 4.184* (30-25)]`

`c_1=1.31J/g^0C`

Therefore, the specific heat capacity of metal is
`1.31J/g^0C`