Question

Find the heat required to raise the temperature of 20 mL of water from $ \text{100}{{\text{ }}^{\text{0}}}\text{C} $ to $ \text{500}{{\text{ }}^{\text{0}}}\text{C} $.

Answer 1

Hint: The amount of heat required to raise the temperature of unit mass of a substance through $ \text{1}{{\text{ }}^{\text{0}}}\text{C} $ is called the Specific Heat Capacity of the substance, denoted by the symbol “c”. The specific heat can have two value, one is the specific heat capacity at constant pressure ( $ \text{Cp} $ ) and the other is the specific heat capacity at constant volume ( $ \text{Cv} $ ). We can calculate the heat required using the formula given.

Formula Used: $ q = m \times {C_v} \times \Delta T $

Complete step by step solution:
To solve this question, we need to know three things, the mass of the substance m, the change in temperature $ \text{ }\Delta\text{ T} $ , and the specific heat capacity of the substance.
Here the volume of water is given which is equal to 20 mL
The density of water = $ \text{1 g/c}{{\text{m}}^{\text{3}}} $
Therefore the mass of water = volume of water multiplies by the density of water = 20 g.
The temperature change, $ \text{ }\Delta\text{ T} $ = $ {{\left( 500-100 \right)}^{0}}\text{C} $ = $ \text{400}{{\text{ }}^{\text{0}}}\text{C} $ = $ 400+273=673\text{K} $
The specific heat capacity of water = $ \text{4}\text{.18 J/g}{{\text{/}}^{\text{0}}}\text{C} $
Therefore the amount of heat required to raise the temperature of 20 grams of water through $ \text{400}{{\text{ }}^{\text{0}}}\text{C} $ = $ 20\times 400\times 4.18=33400 $ joules or $ \text{33}\text{.4 kJ} $
Therefore, $ \text{33}\text{.4 kJ} $ of heat is required to raise the temperature of 20 mL of water from $ \text{100}{{\text{ }}^{\text{0}}}\text{C} $ to $ \text{500}{{\text{ }}^{\text{0}}}\text{C} $ .

Note:
The specific heat capacities of different substances is different and that is why it is specific for a particular substance. For example the specific heat capacity of solid aluminium is $ \text{0}\text{.904 J/g}{{\text{/}}^{\text{0}}}\text{C} $ and that of solid iron is $ \text{0}\text{.449 J/g}{{\text{/}}^{\text{0}}}\text{C} $ . This means that more heat will be required to raise the temperature of unit mass of aluminium through $ \text{1}{{\text{ }}^{\text{0}}}\text{C} $ than that required for the same mass of iron to raise the temperature through the same range.