Question

How much heat is needed to heat 45 kg of water from 18°C to 83°C?

Answer 1

Answer:

The Old Thermodynamist says:

Recall ΔH = m•Cp•ΔT where H is enthalpy (heat), m is mass Cp is heat capacity and ΔT is temperature change.

The heat capacity of water is dependent on temperature, but is commonly taken as 4.186kJ/kg-K

In this case, we have 8kg of water and a temperature change ΔT = 80K

So ΔH = 8(4.186)(80) = 2679.0kJ

The heat required is 2679.0kJ (which is 0.744kWh

Explanation:

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Answer 2

Required Heat Energy can be found out by following formula :

Q= m × c × (Temperature difference )

Where ,

m=mass of substance

C=Specific heat of substance ,

here it is 4.18kj/kg-c or (4.18kj/kg-k) for water

So ,Q= 8 × 4.18 × (90–10)

or,Q= 2675.2 kj. ………………..(Ans)

The Old Thermodynamist says:

Recall ΔH = m•Cp•ΔT where H is enthalpy (heat), m is mass Cp is heat capacity and ΔT is temperature change.

The heat capacity of water is dependent on temperature, but is commonly taken as 4.186kJ/kg-K

In this case, we have 8kg of water and a temperature change ΔT = 80K

So ΔH = 8(4.186)(80) = 2679.0kJ

The heat required is 2679.0kJ (which is 0.744kWh)
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