Question

To make yourself some coffee, you put one cup of water (246 gg ) in a small pot on the stove. Part A What quantity of energy must be transferred thermally to the water to raise its temperature from 20 ∘C∘C to 100 ∘C∘C?

Answer 1

Answer:

The required heat energy is $\bf{19680~cal}$.

Explanation:

The heat energy 'Q' required to raise the temperature of water is given by

$Q = M~s~\Delta \theta$

where 'M' is the mass of water, 's' is the specific heat capacity of water and '$\Delta \theta$' is the change of temperature.

Given, $M = 246 gm, \Delta \theta = (100^{0}~C - 20^{0}~C) = 80^{0}~C$ and we know that the specific heat capacity of water is $s = 1~cal~gm^{-1}~^{0}C^{-1}$.

Substituting the values in the above expression, the required heat energy is

$Q = 246~gm \times 1~cal~gm^{-1}~^{0}C^{-1} \times 80^{0}~C = 19680~Cal$