
A metal drill of power \[40W\] , drills a hole in a lead cube of mass \[1\,kg\] in \[8.5\,s\]. The specific heat capacity of lead is \[130\,Jk{g^{ - 1}}^\circ {C^{ - 1}}\]. If all the mechanical energy of the drill changes into heat energy, calculate the rise in temperature of the lead cube.
Answer
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Hint: We can find the change in temperature by using the formula of specific heat capacity. We can find the value of energy and use this value as the heat produced as all of the mechanical energy is converted.
Formulas used:
The energy of the metal drill is given by \[E = Pt\].
The value of heat produced can be found using the formula,
\[H = mC\Delta T\]
Where \[m\] is the mass of the lead cube, \[C\] is the specific heat capacity of the drill, \[\Delta T\] is the change in temperature, \[P\] is the power of the drill and \[t\] is the time taken to drill through.
Complete step by step answer:
Let us start by gathering the given information, the power of the metal drill is, \[P = 30\,W\].Mass of the lead cube is, \[m = 1\,kg\]. Time taken for the metal drill to drill a hole in the cube, \[t = 8.5s\]. Specific heat capacity of lead is, \[C = 130Jk{g^{ - 1}}^\circ {C^{ - 1}}\]
Energy is given by the formula, \[E = Pt\].We substitute the values and arrive at,
\[E = Pt \\
\Rightarrow E= 40 \times 8.5 \\
\Rightarrow E= 340\,J\]
It is said that all this mechanical energy will be converted to heat energy so, we conclude that the heat produced will be the same value. We now apply this value on the formula \[H = mC\Delta T\] to find the change in temperature.
\[\Delta T = \dfrac{H}{{mC}} \\
\Rightarrow \Delta T= \dfrac{{340}}{{1 \times 130}} \\
\therefore \Delta T = 2.615^\circ C\]
Therefore, the rise in temperature will be \[2.615^\circ C\].
Note: Specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of unit mass of the given substance by one degree. Substances having a small specific heat capacity are very useful as material in cooking instruments such as frying pans, pots, kettles and so on, because, when a small amount of heat is applied it will heat quickly.
Formulas used:
The energy of the metal drill is given by \[E = Pt\].
The value of heat produced can be found using the formula,
\[H = mC\Delta T\]
Where \[m\] is the mass of the lead cube, \[C\] is the specific heat capacity of the drill, \[\Delta T\] is the change in temperature, \[P\] is the power of the drill and \[t\] is the time taken to drill through.
Complete step by step answer:
Let us start by gathering the given information, the power of the metal drill is, \[P = 30\,W\].Mass of the lead cube is, \[m = 1\,kg\]. Time taken for the metal drill to drill a hole in the cube, \[t = 8.5s\]. Specific heat capacity of lead is, \[C = 130Jk{g^{ - 1}}^\circ {C^{ - 1}}\]
Energy is given by the formula, \[E = Pt\].We substitute the values and arrive at,
\[E = Pt \\
\Rightarrow E= 40 \times 8.5 \\
\Rightarrow E= 340\,J\]
It is said that all this mechanical energy will be converted to heat energy so, we conclude that the heat produced will be the same value. We now apply this value on the formula \[H = mC\Delta T\] to find the change in temperature.
\[\Delta T = \dfrac{H}{{mC}} \\
\Rightarrow \Delta T= \dfrac{{340}}{{1 \times 130}} \\
\therefore \Delta T = 2.615^\circ C\]
Therefore, the rise in temperature will be \[2.615^\circ C\].
Note: Specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of unit mass of the given substance by one degree. Substances having a small specific heat capacity are very useful as material in cooking instruments such as frying pans, pots, kettles and so on, because, when a small amount of heat is applied it will heat quickly.
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