Question

An object with a mass of `110 g` is dropped into `750 mL` of water at `0^@C`. If the object cools by `70 ^@C` and the water warms by `3 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`C_"object"=1.22"J/g°C"`

Explanation:

This question is based on the concept that `"heat lost"="heat gained"`. That is, all the heat lost by the object is gained by the water (`q_"water"=q_"object"`).

The equation to find the change in heat is `q=m*C*DeltaT`, and `C_"water"=4.184"J/g°C"`.

Since no density or mass for water is given, it can be assumed that `rho_"water"=1"g/mL"` so there are `750"g"` of water.

Using the change in heat equation, `q_"water"=m_"water"*C_"water"*DeltaT_"water"`
`q_"object"=m_"object"*C_"object"*DeltaT_"object"`

Since `q_"water"=q_"object"`, these two equations can be set equal to each other.
`m_"water"*C_"water"*DeltaT_"water"=m_"object"*C_"object"*DeltaT_"object"`

Solving this for the desired variable,
`C_"object"=(m_"water"*C_"water"*DeltaT_"water")/(m_"object"*DeltaT_"object")`

And plugging in the numbers
`C_"object"=((750"g")(4.184"J/g°C")(3"°C"))/((110"g")(70"°C"))`

Using a calculator
`C_"object"=1.22"J/g°C"`