Question

Calculate the heat required to raise the temperature of 0.75 kg of water from $ {5^ \circ }C $ to $ {90^ \circ }C $ ?

Answer 1

Hint: To solve this problem we need to know the specific capacity of water. It is denoted by c. Specific capacity is the heat required to raise the temperature by $ {1^ \circ }C $ for 1 g of water. Specific heat capacity has a SI unit of Joule per kelvin (J/K) and is an extensive property i.e. it doesn’t depend on the amount of the material present.

Complete answer:
To know the amount of heat required to raise the temperature, we’ll use the relationship between Heat, temperature, and mass of the substance. The formula to be used is: $ Q = mc\Delta T $ -- (1)
Where m is the mass of the substance, Q is the heat required, c is the specific capacity and $ \Delta T $ is the change in the temperature.
First, we’ll find the specific capacity of water, which is the heat required to raise the temperature of 1g of water by $ {1^ \circ }C $ . The amount of heat Q will be numerically equal to c
Since $ m = 1g,\Delta T = {1^ \circ }C $ the heat required will be equal to 1 cal. On converting it into joules we get the specific capacity as $ 1cal/g{/^ \circ }C = 4.186J/g{/^ \circ }C $
The values given to us is: $ m = 0.75kg = 750g,c = 4.186J/g{/^ \circ }C,\Delta T = {(90 - 5)^ \circ }C = {85^ \circ }C $
Substituting the values in equation (1), we get $ Q = 750 \times 4.186 \times 85 = 266858J $
Converting the energy in Joules to Kilojoules, $ 1kJ = 1000J $
Therefore, energy $ Q = 266858J = 266.858kJ $
The answer we require is $ 266.858kJ $ of heat.

Note:
We could have also used the heat capacity, which is the heat required to raise the temperature by 1 degree, but it would have been a longer route. Specific heat and heat capacity are different terms so avoid confusing them. We had got the specific heat capacity in the CGS unit $ cal/g{/^ \circ }C $ , the MKS unit for specific heat capacity is $ J/kg/K $ . The conversion factor that needs to be remembered is: $ 1cal/g{/^ \circ }C = 4200J/kg{/^ \circ }C $
The temperature can be converted from degrees to kelvin by using the relation: $ T(K) = T{(^ \circ }C) + 273 $ .