Question

Calculate heat required to raise the temperature of 1 g of water by 1$^{\circ}$ C ?

Answer 1

Hint: We need to know the specific heat of water in order for us to determine the amount of heat required to raise the temperature of a given mass of water by 1 degree as can be seen in the formula.

Formula used:
Specific heat for a substance is given by:
$c = \dfrac{\Delta Q}{M \Delta T}$
where Q is used to denote heat, T is for temperature and M is for mass.

Complete answer:
The specific heat of a substance is defined as the amount of heat required to raise the temperature of unit mass of the substance through one degree. As we are given 1 g of mass of water and the temperature has to be raised by 1$^{\circ}$ C, we will get the amount of heat to be numerically equal to specific heat:
$\implies c = \Delta Q$.
Now, in c.g.s. units, the specific heat of water is c = 1 cal $g^{-1 \circ} C^{-1}$. Therefore the heat required will be 1 calorie.
Therefore, the heat required to raise the temperature of 1 g of water by 1$^{\circ}$ C is 1 cal.
In MKS units, 1 calorie = 4.186 joule.

Additional information:
The MKS unit for specific heat is J $kg^{-1} K^{-1}$ whereas c.g.s. unit is cal $g^{-1 \circ} C^{-1}$. To make a conversion of specific heat from c.g.s. to MKS, one must first use the relation:
$1 cal g^{-1 \circ} C^{-1} = 4200 J kg^{-1 \circ} C^{-1}$.
Then one must convert the Celsius temperature to Kelvins by using:
T (in K) = t( in $^{\circ}$ C) + 273.
273 K marks zero degree temperature in Celsius scale.

Note:
Here we could have used the definition of heat capacity or thermal capacity instead of specific heat too. By definition the heat capacity of a body is defined as the amount of heat required to raise the temperature of the body by 1 degree. But the answer would have taken a longer route then. Therefore one should not confuse specific heat with heat capacity for the question. The answer simply lies in the definition of specific heat itself. If one remembers that, it could save a great deal of time.