Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A steam boiler made up of steel weighs The boiler contains <img alt="400 \, \, \mat

A steam boiler made up of steel weighs 900 \mathrm{~kg}. The boiler contains 400 \, \, \mathrm{kg} of water. Assuming 70 \% of the heat is delivered to the boiler and water, how much heat is required to raise the temperature of the whole from \mathrm{10^{\circ} \mathrm{C} \, \, to \, \, 100^{\circ} \mathrm{C}} ? The heat capacity of steel is \mathrm{0.11 \mathrm{kcal} / \mathrm{kg}-\mathrm{K}} and the heat capacity of water is \mathrm{1 \mathrm{kcal} / \mathrm{kg} - \mathrm{K}.}

Option: 1

31437 kcal


Option: 2

31400 kcal


Option: 3

31437 kcal


Option: 4

29845 kcal


Answers (1)

best_answer

Given data:

\mathrm{Weight \, \, of \, \, steel }  \mathrm{\left(m_{\text {steel }}\right): 900 \mathrm{~kg}}

\mathrm{\text { Weight of water }\left(m_{\text {water }}\right): 400 \mathrm{~kg}}

\mathrm{\text { Heat capacity of steel }\left(C_{\text {steel }}\right): 0.11 \mathrm{kcal} / \mathrm{kg}-\mathrm{K}}

\mathrm{\text { Heat capacity of water }\left(C_{\text {water }}\right): 1 \mathrm{kcal} / \mathrm{kg} \text {-K }}

\mathrm{\text { Initial temperature }\left(T_{\text {initial }}\right): 10^{\circ} \mathrm{C}}

\mathrm{\text { Final temperature }\left(T_{\text {final }}\right): 100^{\circ} \mathrm{C}}

\mathrm{\text { Efficiency }(\eta): 70 \%}
We need to calculate the heat (Q) required to raise the temperature of the whole system from \mathrm{10^{\circ} \mathrm{C}} to \mathrm{100^{\circ} \mathrm{C}.}

The heat transferred to the system can be calculated using the formula:

  \mathrm{ Q=\eta \cdot\left(m_{\text {steel }} \cdot C_{\text {steel }}+m_{\text {water }} \cdot C_{\text {water }}\right) \cdot \Delta T }

where \mathrm{ \Delta T} is the change in temperature \mathrm{ \left(T_{\text {final }}-T_{\text {initial }}\right).}

Substituting the given values:

\mathrm{ Q=0.70 \cdot((900 \mathrm{~kg} \cdot 0.11 \mathrm{kcal} / \mathrm{kg}-\mathrm{K})+(400 \mathrm{~kg} \cdot 1 \mathrm{kcal} / \mathrm{kg}-\mathrm{K})) \cdot\left(100^{\circ} \mathrm{C}-10^{\circ} \mathrm{C}\right) }
Solving for Q :

\mathrm{ Q=(0.70) \cdot(99 \mathrm{kcal} / \mathrm{K}+400 \mathrm{kcal} / \mathrm{K}) \cdot\left(90^{\circ} \mathrm{K}\right)=31437 \mathrm{kcal} }

Therefore, the heat required to raise the temperature of the whole system from \mathrm{10^{\circ} \mathrm{C} \, \, to \, \, 100^{\circ} \mathrm{C} \, \, is \, \, 31437 \, \, \mathrm{kcal}.}

So, correct option is (3)

Posted by

Info Expert 30

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question