Question

1. what is the latent heat of fusion for water? *

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144 btu/lbm
970 btu/lbm
540 cal/g
80 cal/g

1. using ( q=m c delta t ), if 5 lbm of water is heated from ( 60^{circ} mathrm{f} ) to ( 180^{circ} mathrm{f} ), and ( c_{1}=1.0 ) * 1 point

btu/lbm ( { }^{circ} mathrm{f} ), what is ( q ) ?
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240 btu
900 btu
600 btu
120 btu
the specific heat of aluminum is ( 0.22 mathrm{btu} / mathrm{lbm} cdot{ }^{circ} mathrm{f} ). how much energy is
needed to heat 10 lbm of aluminum by ( 50^{circ} mathrm{f} ) ?
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500 btu
110 btu
220 btu
1100 btu

1. which material typically has the highest specific heat capacity? *

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steel
aluminum
copper
water

1. if a substance has a density of ( 62.4 mathrm{lbm} / mathrm{ft}^{3} ), what is its specific volume? *

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0.032 ( mathrm{ft}^{3} / mathrm{lbm} )
0.016 ( mathrm{ft}^{3} / mathrm{lbm} )
62.4 ( mathrm{ft}^{3} / mathrm{lbm} )
124.8 ( mathrm{ft}^{3} / mathrm{lbm} )

1. the relationship between density (p) and specific volume (v) is *

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( p=2 v )
( p=1 / v )
( p=v )
( p=v^{2} )

Explanation:

Step1: Identify the formula and variables

The formula for heat energy is \( Q = mc\Delta T\). Here, \(m = 10\ \text{lbm}\), \(c=0.22\ \text{Btu/lbm}\cdot^{\circ}\text{F}\), and \(\Delta T = 50^{\circ}\text{F}\)

Step2: Substitute the values into the formula

Substitute \(m = 10\), \(c = 0.22\), and \(\Delta T=50\) into \(Q = mc\Delta T\). So, \(Q=(10)(0.22)(50)\)

Step3: Calculate the result

First, calculate \((10)(0.22)=2.2\), then \(2.2\times50 = 110\ \text{Btu}\)

Answer:

110 Btu