Question

A beaker containing 400g of water has 1200J of work done on it by stirring and 200cal of heat added to it from a hot plate.8.What is the temperature change of the water?A.1.2°C B.2.1°C C.1.4°C D.4.1°C

Answer 1

A.1.2°C

Step-by-step explanation

The First Law of Thermodynamics states that heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy.

To find internal energy, you have to add the heat added in the system and work done in the system because the work done is not lost but rather it is added in the system.

`\begin{gathered} \Delta U=q+W \\ where\Delta U\text{ is the change in the internal energy} \\ q\text{ the heat added to the system} \\ W\text{ is the work done by the systeme} \end{gathered}`

Step 1

a)let

`q=200\text{ cal}`

to add the energy it must have the same measure unit, so let's convert calories into Julies

remember that

`\begin{gathered} 1\text{ cal}\Rightarrow4.184\text{ J} \\ so \\ 200\text{ cal}\Rightarrow200(4.814\text{ J})\Rightarrow836.8\text{ J} \end{gathered}`

b) now, replace in the formula

`\begin{gathered} \Delta U= q+W \\ \Delta U=836.8\text{ J+1200 J} \\ \Delta U=2036.8\text{ J} \end{gathered}`

now, we have the change of internal energy

Step 2

now, let's find the change in temperature

Use the calorimetry formula.

`\begin{gathered} Q=mc∆T \\ \end{gathered}`

where m is the mass, Q = heat energy, c = specific heat capacity, and ∆T = change in temperature

a)

let

`mass=\text{ 0.4 kg}`

now,

`\begin{gathered} Q=mC\Delta T \\ 2036.8\text{ j=0.4 kg*4184 }(J)/(Kg)|C*\Delta T \\ 2036.8=1673.6\Delta T \\ divide\text{ both sides by 1673.6} \\ (2,036.8)/(1673.6)=\frac{1,673.6\text{ }\Delta T}{1673.6} \\ 1.21=\Delta T \\ rounded \\ \Delta T=1.2\text{ \degree C} \end{gathered}`

so, the answer is

A.1.2°C