Question

How much heat (kJ) is needed to raise the temperature of $100.0$ grams of water from ${25.0^0}C$ to ${50.0^0}C$ ?

Answer 1

This question is related to thermodynamics. For solving this question, we need to know a term called specific heat capacity which is defined as the heat required to raise the temperature of one gram of substance to one degree Celsius.

Complete step by step answer: Here the substance is water. So we need to know the value of specific heat capacity of water to solve this question. The specific heat capacity of water is $4.18J/g{}^0C$ . This is the heat required to increase the temperature by one degree Celsius. To increase the temperature by ${2^0}C$ , the heat required will be $2 \times 4.18J$ . In the same way, $4.18J$ is the heat required to raise the temperature of one gram of water. Hence to increase the temperature of $2g$ water by one degree Celsius, heat required will be $2 \times 4.18J$ . From this concept we can derive an equation as follows.
The heat required to raise the temperature of n grams of a substance by m $^0C$ will be,
Heat = specific heat capacity of the substance $\times$ mass of the substance in grams $\times$ temperature difference
Let us rewrite the equation as,
$q = mc\Delta T$
Where q is the heat required, m is the mass of the substance, c is the specific heat capacity of the substance and $\Delta T$ is the temperature difference.
Given that,
$m = 100g$
$\Delta T = {50^0}C - {25^0}C = {25^0}C$
$c = 4.18J/g{}^0C$
Let us substitute the values.
$q = 100 \times 25 \times 4.18 = 10450J$
Hence the heat required is $10450J$ . Converting it to kJ, the heat required to raise the temperature of $100.0$ grams of water from ${25.0^0}C$ to ${50.0^0}C$ is $10.45kJ$ .

Note:
Water requires more heat to raise its temperature compared to other common substances. The temperature can be taken in the unit of Kelvin also. Because the difference in temperature will be the same whether it is expressed in ${}^0C$ or Kelvin.