Question

A silver block, initially at 57.2 ∘C, is submerged into 100.0 g of water at 24.8 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 26.5 ∘C. Part A What is the mass of the silver block? Express your answer to two significant figures and include the appropriate units. mm =

Answer 1

96.5 grams is the mass of the silver block.

Step-by-step explanation:

Heat lost by iron will be equal to heat gained by the water

`-Q_1=Q_2`

Mass of iron =
`m_1=?`

Specific heat capacity of silver=
`c_1=0.240 J/g^oC`

Initial temperature of the silver=
`T_1=120^oC`

Final temperature =
`T_2=T=26.5^oC`

`Q_1=m_1c_1* (T-T_1)`

Mass of water=
`m_2=100.0 g`

Specific heat capacity of water=
`c_2=4.184 J/g^oC`

Initial temperature of the water =
`T_3=24.8 ^oC`

Final temperature of water =
`T_2=T=26.5^oC`

`Q_2=m_2c_2* (T-T_3)`

`-Q_1=Q_2`

`-(m_1c_1* (T-T_1))=m_2c_2* (T-T_3)`

On substituting all values:

`-(m_1* 0.240 J/g^oC* (26.5^oC-57.2^oC))=100.0 g* 4.184 J/g^oC* (26.5^oC-24.8^oC)`

we get,
`m_1 = 96.5 g`

96.5 grams is the mass of the silver block.