Question

In a large building, oil is used in a steam boiler heating system. The combustion of 1.0 lb of oil provides 2.2×107J. Part A How many kilograms of oil are needed to heat 110 kg of water from 25 ∘C to 100∘C? (Assume that 1lb=454g, the specific heat of liquid water is 4.184 J/g∘C.)

Answer 1

0.712 kg

Step-by-step explanation:

Data given:

Heat provide by 1 lb Oil = 2.2×107J

mass of water = 110 kg

Convert mass from Kilogram to gram

1 kg = 1000 g

110 kg = 1000 x 110 = 110 x 10³ g

initial temperature = 25 °C

final temperature = 100 °C

specific heat of liquid water = 4.184 J/g∘C

Solution:

first we will find amount of heat required by given amount water

Formula will be used

Q = Cs.m.ΔT . . . . . . . .(1)

Where:

Q = amount of heat

Cs = specific heat of water = 4.184 J/g °C

m = mass

ΔT = (t2 - t1) = Change in temperature

First we have to find ΔT

ΔT = (t2 - t1)

ΔT = (100 °C - 25 °C )

ΔT = 75 °C

Put values in above equation 1

Q = (4.184 J/g °C)(110 x 10³ g)(75 °C)

Q = 34.518 x 10⁶ J

So,

we come to know that 34.518 x 10⁶ J heat is required to heat 110 kg

So,

if 1.0 lb of oil provides 2.2×10⁷ J then how many lb of oil will produces 34.518 x 10⁶ J heat.

Apply unity formula

1.0 lb of oil ≅ 2.2×10⁷ J of heat

X lb of oil ≅ 34.518 x 10⁶ J of heat

Do cross multiplication

X lb of oil = 1.0 lb x 34.518 x 10⁶ J / 2.2×10⁷ J

X lb of oil = 1.569 lb

So, 1.569 lb of oil will produce 34.518 x 10⁶ J of heat required for 110 kg of water.

Now convert lb to grams

as data given

1 lb = 454 g

So,

1.569 lb = 1.569 x 454 = 712.33 g

As we want to know the amount of oil in Kg so convert grams to kg.

1 g = 0.001 Kg

so,

712.33 g = 712.33 x 0.001 = 0.712 kg

So,

kilograms of oil are needed = 0.712 kg

0.712 kg kilograms of oil are needed to heat 110 kg of water from 25 ∘C to 100∘C.