Question

The heat of combustion of bituminous coal is 2.50 x 10⁴ J/g. What quantity of the coal is required to produce the energy to convert 106.9 pounds of ice at 0.00°C to steam at 100.°C?specific heat (ice) = 2.10 J/g°Cspecific heat (water) = 4.18 J/g°Cheat of fusion = 333 J/gheat of vaporization = 2258 J/g

Answer 1

mc=5.84kg

Step-by-step explanation:

Given

m i = 106.9 l b s

m i = 48480.7 g

T 1 = T i = 0 °C

T 2 = T s = 100 °C

C p i = 2.10 J / g − C

C p w = 4.18 J / g − C

H f = 333 J / g

H v = 2258 J / g

H c = 25000.00 J / g

According to the first law of thermodynamics, the heat supplied by the coal must be equal to the heat required to convert the ice into water and then the water into steam:

Q c = Q i + Q w + Q s . . . . . . . . . . . . . . . . . . .(eq:1)

Since the water will undergo a change in temperature (0 C to 100 C )

Q i = m i ∗ H f..........(eq:2)

Q s= m s ∗ H v..........(eq:3)

heat allows for the change in temperature, calculated as:

Q w = m w ∗ C p w ∗ ( T 2 − T 1 )...........(eq:4)

Integrating Equations 2, 3 and 4 into Equation 1, we get:

Q c = m i ∗ H f + m w ∗ C p w ∗ ( T 2 − T 1 ) + m s ∗ H v ...........(eq:5)

m i = m w = m s = m

Q c = m ∗ ( H f + C p w ∗ ( T 2 − T 1 ) + H v )

Q c = 48480.7 ∗ ( 333 + 4.18 ∗ ( 100 − 0 ) + 2258 ) J

Q c = 145878426.3 J

The heat of the coal is the product of the mass of the coal and its heat of combustion.

Thus:

m c = Q c / H c

m c = 145878426.3/ 2.5 x 10 4 g

m c = 5835.137052 g

mc=5.84kg