Question

If the metal from problem 4 was initially at room temperature (22 0 C), what would the final temperature of the metal be? You know that you add 120 joules of energy to the metal. What change in temperature would you observe Q = is energy as Heat, 120 Joulesm = mass in grams, 5.0 gramsc = is the specific heat capacity, 0.385 J/g°CΔT = the change in temperature, calculated as Final Temperature - Initial T120 = 5 * 0.385 * T120 = 1.925TT = 62°C of change in temperature

Answer 1

The question is mostly solved. The definition of heat is used for this problem which tells us:

$Q=mCp\Delta T$

Where,

Q is the heat added to the system, 120 J

m is the mass of the metal, 5.0 g

Cp is the specific heat of the metal, 0.385J/g°C

dT is the change of temperature:

$\Delta T=T_2-T_1$

T2 is the final temperature, unknown

T1 is the initial temperature, 22°C

We clear the final temperature from the equation:

$\begin{gathered} Q=mCp(T_2-T_1) \\ Q=mCpT_2-mCpT_1 \\ T_2=\frac{Q+mCpT_1}{mCp} \end{gathered}$

Now, we replace the known data:

$T_2=\frac{120J+5.0g\times0.385\frac{J}{g\degree C}\times22\degree C}{5.0g\times0.385\frac{J}{g\degree C}}$$\begin{gathered} T_2=\frac{120+5.0\times0.385\times22}{5.0\times0.385}\degree C \\ T_2=84\degree C \end{gathered}$

Answer:

The final temperature of the metal will be 84°C

The change in the temperature will be 84°C-22°C=62°C