Question

The specific heat capacity of liquid water is 4.18 kJ/g C, how would you calculate the quantity of energy required to heat 1.00 g of water from 26.5 C to 83.7 C?

Answer 1

${s_{water}} = 4.18\dfrac{J}{{g^ \circ C}}$
If Q is the heat supply to a substance to mass ‘m’ such that its temperature by $\Delta T$ , then the relation between Q, m and $\Delta T$ is given as
$Q = ms\;\Delta T\;\;\;\;$

Complete step by step answer:
If Q is the heat supply to a substance to mass ‘m’ such that its temperature by $\Delta T$ , then the relation between Q, m and $\Delta T$ is given as
$Q = ms\;\Delta T\;\;\;\;$where S is the specific heat capacity of the material
Alternatively ‘spc’ is also defined as the amount of heat required to raise the temperature of 1kg of a substance by ${1^o}C$.
S=$\dfrac{Q}{{m\Delta T}}$
If m=1, $\Delta T\; = {1^o}C$
Then, S=Q , after that spc changes
S.I. unit = Joule $\mathop K\nolimits^{ - 1}$/kg
$q = 1.00g \times 4.18\dfrac{J}{{g{}^0C}} \times (83.7 - 26.5){}^0C$
q=239.096 J

Hence the final answer is q = 239.096

Additional Information:
It is to be noted that the specific heat capacity of a substance is the amount of energy required to raise the temperature of 1 kg of the substance by one degree Celsius. Always remember that, the specific heat capacity of water is ${\text{42}}00{\text{ J}}/{\text{kg}}\,^\circ {\text{C}}$.

Note: Always remember that high specific heat capacity means the substance takes a lot of heat to raise the temperature (takes a long time to heat and cool). Low specific heat capacity means only a little amount of heat is needed to raise the temperature (takes only a little time to heat and cool). Water has the highest specific heat capacities and metals have lowest specific heat capacities.