A steam boiler made up of steel weighs 900 kg. The boiler contains 400 kg of water. Assuming 70% of the heat is delivered to boiler and water, how much heat is required to raise the temperature of the whole from `10^(@) C to 100^(@)C ?` Heat capacity of steel is 0.11 kcal `kg ^(-1)K^(-1)` and heat capacity of water is 1 kcal `kg^(-1) K^(-1)`

Options

A. 65321 kcal
B. 64157 kcal
C. 6869 kcal
D. 66549 kcal

Explanation

By using the equation `Delta H = m _("boller") C _("boller") Delta T + m _("water") C _("water") Delta T` first find out the `Delta H` required for complete heating and then by using the value of `Delta H` calculate the actual heat required when `70%` is transferred to the system `Delta H` required for heating `= m _("boiler") C _("boiler") Delta T = m _("water") C _("water") Delta T`
`= 900 xx 0.1 1 xx 90 + 400 xx 1 xx 90 = 44910 kcal.` Since, only `70%` of heat given is used up to do so.
Thus actual heat required `= ( 44910 xx 100)/( 70) = 64157 kcal`