Question

What is the final temperature if a metal at `75^@ "C"` is dropped into `"185 g"` of water starting at `25^@ "C"`? The heat capacity of the metal is `"55 J/g"*""^@ "C"`.

Answer 1

This is one of those questions where solving it in general would help you more with not messing up on the math.

`T_(f) = [m_wC_wT_(iw) + c_mT_(im)]/(c_m + m_wC_w)`

I get `30.81^@ "C"`, using `C_w = "4.184 J/g"^@ "C"` and `c_m = "55 J/"^@ "C"` for the metal. Here we have `T_(iw) = 25^@ "C"` and `T_(im) = 75^@ "C"`.

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You know that there is conservation of energy, so

`q_m + q_w = 0`,

`q_m = -q_w`

where `m` is metal and `w` is water, while `q` is heat flow in `"J"`. That means you can equate their equations:

`q_m = m_mC_mDeltaT_m`
`q_w = m_wC_wDeltaT_w`

where `m` is mass in `"g"`, `C` is specific heat capacity in `"J/g"^@ "C"` or `"J/g"cdot"K"`, and `DeltaT` is the change in temperature in either `""^@"C"` or `"K"`.

`m_mC_mDeltaT_m = -m_wC_wDeltaT_w`

The metal is hotter, so dropping it into water must cool it down. So, `T_f > T_(im)`, while `T_f < T_(iw)`. That assures you a positive answer in the end:

`m_mC_m(T_f - T_(im)) = -m_wC_w(T_f - T_(iw))`

`m_mC_m(T_f - T_(im)) = m_wC_w(T_(iw) - T_f)`

Distribute the terms:

`m_mC_mT_f - m_mC_mT_(im) = m_wC_wT_(iw) - m_wC_wT_f`

Now get the `T_f` terms to one side and the `T_i` terms to the other side.

`m_mC_mT_f + m_wC_wT_f = m_wC_wT_(iw) + m_mC_mT_(im)`

Finally, factor out `T_f` and divide:

`T_(f)(m_mC_m + m_wC_w) = m_wC_wT_(iw) + m_mC_mT_(im)`

`T_(f) = [m_wC_wT_(iw) + m_mC_mT_(im)]/(m_mC_m + m_wC_w)`

`color(blue)(T_(f) = [m_wC_wT_(iw) + c_mT_(im)]/(c_m + m_wC_w))`

where `m_mC_m = c_m`, the heat capacity in `"J/"^@ "C"`.

So now, we can plug the numbers in and solve for the final temperature.

Answer 2

Let the final temperature be `t^@C`

Heat lost by metal `=55"J/"^@Cxx(75-t)^@C`

`=55(75-t)J`

Heat gained by water
`=185gxx4.2J/(g^@C)xx(t-25)^@C`

`color(blue)("Where "4.2J/(g^@C)" is the sp.heat of water")`

So by calorimetric principle

`185xx4.2xx(t-25)=55xx(75-t)`

Now dividing both sides by 5 we get

`=>37xx4.2xx(t-25)=11(75-t)`

`=>37xx4.2t+11t=11xx75+37xx4.2xx25`

`=>166.4t=4710`

`=>t=4710/166.4~~28.3^@C`