Question

In a large building, oil is used in a steam boiler heating system. The combustion of $1.0lb$ of oil provides $2.4\times {10}^{7}J$. $(3.4,3.6)$

$\left(a\right)$How many kilograms of oil are needed to heat $150kg$ of water from ${22}^{\circ }C$ to ${100}^{\circ }C$ ?

$\left(b\right)$How many kilograms of oil are needed to change $150kg$ of water to steam at ${100}^{\circ }C$ ?

Short answer

Part $\left(a\right)$

$\left(a\right)$$0.93kg$of oil is required to heat $150kg$of water from ${22}^{\circ }C$to ${100}^{\circ }C$.

Part $\left(b\right)$

$\left(b\right)$$6.43kg$of oil is required to heat $150kg$of steam water at${100}^{\circ }C.$

Step-by-step solution

Step: Given information: (Part a) 

Given are:$1.0lb$of oil and role="math" localid="1651854658616" $2.4\times {10}^{7}J$and $150kg$of water and temperature is${22}^{\circ }C-{100}^{\circ }C.$

To find the oil in kilogram.

Step: 2  Finding heat: (Part a) 

The mass of water is$=150\mathrm{kg}$

$=150\times 1000\mathrm{g}$

The change in temperature as${100}^{\circ }\mathrm{C}-{22}^{\circ }\mathrm{C}={78}^{\circ }\mathrm{C}$

role="math" $\text{Heat required}=\text{mass}\times \text{temperature change}\times \text{specific heat}$

role="math" localid="1651854827858" $Heat=150\times 1000\mathrm{g}\times {78}^{\circ }\mathrm{C}\times 4.184\mathrm{J}/{\mathrm{g}}^{\circ }\mathrm{C}$

As aresult,oil in kilogram is$Heat=4.895\times {10}^7\mathrm{J}.$

Step: 3 Finding oil in kilogram: (Part a)

Calculating pounds of oil as

$\text{pounds of oil required}=\frac{\text{Heat required}}{\text{Heat obtained from}1\mathrm{lb}}$

localid="1651855639395" $\text{pounds of oil required}=\frac{4.895\times {10}^7\mathrm{J}}{2.4\times {10}^7\mathrm{J}}$

$\text{pounds of oil required}=2.04lb.$

Where,$1kg=2.04lb$

so,$2.04\mathrm{lb}=\frac{2.04}{2.20}\mathrm{kg}$

As a result,oil need in kilogram as$0.93kg.$

Step: 4 Given information: (Part b) 

Given are: $150g$of water and steam is ${100}^{\circ }C.$

To find the oil in kilogram.

Step: 5 Finding oil in kilogram: (Part b)

Mass of water is $150kg$

Heat vapourization of water is $2260kJ/kg$

role="math" localid="1651855871818" $\text{Heat}=\text{Mass}\times \text{heat of vaporization}$

$Heat=150\mathrm{kg}\times 2260\mathrm{kJ}/\mathrm{kg}$

$Heat==3.4\times {10}^5\mathrm{kJ}$

$Poundsofoil=\frac{3.4\times {10}^5\mathrm{kJ}}{2.4\times {10}^4\mathrm{kJ}/\mathrm{lb}}$

$Poundsofoil=1.413\times {10}^1\mathrm{lb}$

$1.413\times {10}^{\mathrm{l}}\mathrm{lb}=\frac{1.413\times {10}^{\mathrm{l}}}{2.20}\mathrm{kg}$

As a result,the oil in kilogram as$6.4\mathrm{kg}.$