# The specific heat capacity of platinum is 0.032 cal/g·K. How to calculate the heat, in joules, necessary to raise the temperature of a sample of platinum weighing 20.0 g from 15.0 Celsius to 65.0 Celsius?

Feb 28, 2015

You have to convert either the specific heat of platinum, or the actual heat required from Cal/g K to J/g K. I"ll convert the specific heat and calculate the heat directly in Joules.

So, in order to get from calories per g Kelvin to Joules per g Kelvin you must use a conversion factor

0.032"cal"/("g" * "K") * "4.184 J"/"1 calorie" = 0.134"J"/("g" * "K")

The relationship between heat and temperature change is expressed mathematically through this equation

$q = m \cdot c \cdot \Delta T$, where

$q$ - the heat supplied;
$m$ - the mass of the sample;
$c$ - the specific heat of, in your case, platinum;
$\Delta T$ - the change in temperature;

Solving for $q$ will get you

$q = \text{20.0 g" * 0.134"J"/("g" * "K") * (338.15 - 288.15)"K" = "+134 J}$

Notice how little energy is needed to raise the temperature of a $\text{20.0-g}$ sample of platinum by 50 degrees; by comparison, you would need approximately 30 times more energy to raise the temperature of this much water by 50 degrees.