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

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: $T(K) = T{(^ \circ }C) + 273$ .