Question

Calculate the heat absorbed by $125g$of water $\left( {SH = 1.00cal{{\left( {{g^0}C} \right)}^{ - 1}}} \right)$ when the temperature rises from ${21^0}C$ to ${35^0}C$ ?

Answer 1

The heat absorbed is equal to the product of the mass of water, specific heat of water which was represented by SH and the difference or change in temperature in Celsius. Specific heat is defined as the amount of heat required by a unit mass of substance through one degree centigrade.
Formula used:
$q = mC\Delta T$
$q$ is heat absorbed
$m$ is mass of water
$C$ is specific heat
$\Delta T$ is change in temperature

Complete answer:
Given that the mass of water is $125g$
The specific heat is defined as the amount of heat required for a unit mass of substance to raise the temperature through one degree centigrade. It was given as $SH = 1.00cal{\left( {{g^0}C} \right)^{ - 1}}$
The initial temperature is given as ${21^0}C$ and the final temperature is given as ${35^0}C$
Thus, the change in temperature which was represented by $\Delta T$ is $\left( {35 - 21} \right) = {14^0}C$
Substitute the value of change in temperature, specific heat of water, and mass of substance in the above formula to get the heat absorbed by water
$q = 125 \times 1 \times 14 = 1750cal$
The amount of heat absorbed is obtained in calories. To convert this value into joules, multiply the value with $4.18$ as one calorie is equal to $4.18J$
$q = 1750 \times 4.18 = 7322J$

Note:
The specific heat is given in terms of calories, grams and Celsius, thus the temperature must be in Celsius, and mass must be present in grams only. When the values substituted in the formula were of these units the heat absorbed will be obtained in calories, which can be converted later into joules.