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Ksju [112]
3 years ago
15

Here, we want to become proficient at changing units so that we can perform calculations as needed. The basic heat transfer equa

tion is q = m*Cp*T, where q is the amount of heat added or subtracted, Cp is the constant pressure heat capacity, and T is the change in temperature (T = T2 – T1). For example, we need to heat 100 lbs of water from 25 °C to 75 °C. How much energy must we add to the water if the constant pressure heat capacity of water is 4.18 J/g K? (Give answer in both kJ and BTU)
Engineering
1 answer:
netineya [11]3 years ago
7 0

Answer:

9500 kJ; 9000 Btu

Explanation:

Data:

m = 100 lb

T₁ = 25 °C

T₂ = 75 °C

Calculations:

1. Energy in kilojoules

ΔT = 75 °C - 25 °C = 50 °C  = 50 K

m = \text{100 lb} \times \dfrac{\text{1 kg}}{\text{2.205 lb}} \times \dfrac{\text{1000 g}}{\text{1 kg}}= 4.54 \times 10^{4}\text{ g}\\\\\begin{array}{rcl}q & = & mC_{\text{p}}\Delta T\\& = & 4.54 \times 10^{4}\text{ g} \times 4.18 \text{ J$\cdot$K$^{-1}$g$^{-1}$} \times 50 \text{ K}\\ & = & 9.5 \times 10^{6}\text{ J}\\ & = & \textbf{9500 kJ}\\\end{array}

2. Energy in British thermal units

\text{Energy} = \text{9500 kJ} \times \dfrac{\text{1 Btu}}{\text{1.055 kJ}} = \text{9000 Btu}

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Answer:

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Explanation:

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Nc = 1.621

putting the above values in equation.

qu = 45.873 KN/mL

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Answer:

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