Question

The specific heat capacity of platinum is \[0.032 {\text{cal/g K}}\]. How to calculate the heat, in joules, necessary to raise the temperature of a sample of platinum weighing $20.0\;{\text{g}}$ from \[{15^ \circ }\] Celsius to \[{65^ \circ }\] Celsius?

Answer 1

Hint: To solve this question, first we have to convert the specific heat required from \[{\text{Cal/g K}}\] to \[{\text{J/g K}}\]. By doing that, we will arrive at the solution in Joules directly. And then, we will use the formula to solve further.

Formula Used:
We will use the following formula to solve this question
\[q = mc\Delta T\]
Where
$q$ is the heat supplied
$m$ is the mass of the sample
$c$ is the specific heat of platinum
$\Delta T$ is the change in temperature

Complete step-by-step answer:So, in order to get from calories per g Kelvin to Joules per g Kelvin you must use a conversion factor
${\text{c}} = 0.032\dfrac{{{\text{cal}}}}{{{\text{g}} {\text{K}}}} \times \dfrac{{4.184\;{\text{J}}}}{{1{\text{ calorie }}}} = 0.134{\text{ J/g K}}$
According to the question, we have the following information given to us
The mass of the sample, \[m = 20 {\text{g}}\]
The specific heat of platinum, \[c = 0.134{\text{ J/g K}}\]
The change in temperature, \[\Delta T = {65^ \circ }{\text{C}} - {15^ \circ }{\text{C}} = 338.15 {\text{K}} - 228.15{\text{ K}} = 50 {\text{K}}\]
Now, we will easily solve this question by putting all the known values in the above given formula to get
\[q = 20 {\text{g}} \times 0.134\dfrac{{\text{J}}}{{{\text{g}} {\text{K}}}} \times 50 {\text{K}}\]
Upon solving the above equation, we get
\[\therefore q = 134{\text{ J}}\]
Hence, the heat produced is \[134{\text{ J}}\]

Additional Information:
Heat, often called thermal energy, is a form of energy. Energy can be transformed from one form to another (a blender transforms electricity into mechanical energy), but it can neither be generated nor destroyed; energy is conserved instead. The higher the temperature of a material in basic thermodynamics, the more thermal energy it possesses. Furthermore, the more of a given substance, the more total thermal energy the material will possess at a given temperature.

Note:Specific heat values can be determined as follows: when two materials are placed in contact with each other, each initially at a different temperature, heat always flows from the warmer material into the colder material until both materials reach the same temperature. The heat gained by the initially colder material must equal the heat lost by the initially warmer material, according to the law of energy conservation.