Question

A silver block, initially at 56.1 ∘C, is submerged into 100.0 g of water at 24.0 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 28.0∘C. What is the mass of the silver block?

Answer 1

Answer: The mass of the silver block is 256 grams

Step-by-step explanation:

`heat_(absorbed)=heat_(released)`

As we know that,

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

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]` .................(1)

where,

q = heat absorbed or released

`m_1` = mass of silver = ?

`m_2` = mass of water = 100.0 g

`T_(final)` = final temperature =
`28.0^0C`

`T_1` = temperature of silver =
`56.1^oC`

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

`c_1` = specific heat of silver =
`0.233J/g^0C`

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

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

`-m_1* 0.233* (28.0-56.1)=[100.0* 4.184* (28.0-24.0)]`

`m_1=256g`

Therefore, the mass of the silver block is 256 grams