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

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.