Question

Solve the following problem step by step: how much heat is need to convert 0.105 kg of ice at −10

∘
C to steam at 140
∘
C. The specific heat of ice is 2090 J/(kg⋅
∘
C), the latent heat of fusion of water is 3.34×10
5
J/kg, the specific heat of water is 4186 J/(kg
∘
C), the latent heat of vaporization of water is 2.26×10
6
J/kg, and the specific heat of steam is 1520 J/(kg⋅
∘
C). Problem 8: How much heat is needed to increase the temperature of 0.015 kg of ice from −10
∘
C to 0
∘
C ? a. 113.5 J b. 213.5 J c. 313.5 J d. 413.5 J Problem 9: How much heat is needed to convert 0.015 kg of ice at 0
∘
C to 0.015 kg of water at 0
∘
C ? a. 4010 J b. 5010 J c. 6010 J d. 7010 J Problem 10: How much heat is needed to increase the temperature of 0.015 kg of water from 0
∘
C to 100
∘
C ? a. 6279 J b. 7279 J c. 8279 J d. 9279 J Problem 11: How much heat is needed to convert 0.015 kg of water at 100
∘
C to 0.015 kg of steam at 100
∘
C ? a. 13900 J b. 23900 J c. 33900 J d. 43900 J Problem 12: How much heat is needed to convert 0.015 kg of steam at 100
∘
C to 0.015 kg of steam at 140
∘
C ? a. 612 J b. 712 J c. 812 J d. 912 J Problem 13: How much heat is need to convert 0.105 kg of ice at −10
∘
C to steam at 140
∘
C ? Use your answers from Problems 8-12. a. 46414.5 J b. 56414.5 J c. 66414.5 J d. 76414.5 J

Answer 1

Problem 8:

Answer: b. 213.5 J

Step-by-step explanation:

Use the formula Q = mcΔT, where Q is heat, m is mass, c is specific heat, and ΔT is change in temperature.

Plug in values: Q = (0.015 kg)(2090 J/kg°C)(10°C) = 213.5 J

Problem 9:

Answer: a. 4010 J

Step-by-step explanation:

Use the formula Q = mL, where Q is heat, m is mass, and L is latent heat of fusion.

Plug in values: Q = (0.015 kg)(3.34 x 10^5 J/kg) = 4010 J

Problem 10:

Answer: c. 8279 J

Step-by-step explanation:

Use the formula Q = mcΔT (same as in Problem 8, but with water's specific heat).

Plug in values: Q = (0.015 kg)(4186 J/kg°C)(100°C) = 8279 J

Problem 11:

Answer: c. 33900 J

Step-by-step explanation:

Use the formula Q = mL, where Q is heat, m is mass, and L is latent heat of vaporization.

Plug in values: Q = (0.015 kg)(2.26 x 10^6 J/kg) = 33900 J

Problem 12:

Answer: d. 912 J

Step-by-step explanation:

Use the formula Q = mcΔT (same as in Problem 8, but with steam's specific heat).

Plug in values: Q = (0.015 kg)(1520 J/kg°C)(40°C) = 912 J

Problem 13:

Answer: d. 76414.5 J

Step-by-step explanation:

Add up the heat amounts from Problems 8-12 to get the total heat required: 213.5 J + 4010 J + 8279 J + 33900 J + 912 J = 76414.5 J

Complete Question:

In Problems 8-13 we will solve the following problem step by step: how much heat is need to convert
`$0.105 \mathrm{~kg}$` of ice at
`$-10^(\circ) \mathrm{C}$` to steam at
`$140^(\circ) \mathrm{C}$`. The specific heat of ice is
`$2090 \mathrm{~J} /\left(\mathrm{kg} \cdot{ }^(\circ) \mathrm{C}\right)$`, the latent heat of fusion of water is
`$3.34 * 10^5 \mathrm{~J} / \mathrm{kg}$`, the specific heat of water is
`$4186 \mathrm{~J} /\left(\mathrm{kg} \cdot{ }^(\circ) \mathrm{C}\right)$`, the latent heat of vaporization of water is
`$2.26 * 10^6 \mathrm{~J} / \mathrm{kg}$`, and the specific heat of steam is
`$1520 \mathrm{~J} /\left(\mathrm{kg} \cdot{ }^(\circ) \mathrm{C}\right)$`.

Problem 8:

How much heat is needed to increase the temperature of
`$0.015 \mathrm{~kg}$` of ice from
`$-10^(\circ) \mathrm{C}$` to
`$0^(\circ) \mathrm{C}$` ?

a. 113.5 J

b. 213.5 J

c. 313.5 J

d. 413.5

Problem 9:

How much heat is needed to convert
`$0.015 \mathrm{~kg}$` of ice at
`$0^* \mathrm{C}$` to
`$0.015 \mathrm{~kg}$` of water at
`$0^* \mathrm{C}$` ?

a. 4010 J

b. 5010 J

c. 6010 J

d. 7010 J

Problem 10:

How much heat is needed to increase the temperature of
`$0.015 \mathrm{~kg}$` of water from
`$0^(\circ) \mathrm{C}$` to
`$100^(\circ) \mathrm{C}$`

a. 6279 J

b. 7279 J

c. 8279 J

d. 9279 J

Problem 11:

How much heat is needed to convert
`$0.015 \mathrm{~kg}$` of water at
`$100^(\circ) \mathrm{C}$` to
`$0.015 \mathrm{~kg}$` of steam at
`$100^(\circ) \mathrm{C}$` ?

a. 13900 J

b. 23900 J

c. 33900 J

d. 43900 J

Problem 12:

How much heat is needed to convert
`$0.015 \mathrm{~kg}$` of steam at
`$100^(\circ) \mathrm{C}$` to
`$0.015 \mathrm{~kg}$` of steam at
`$140^(\circ) \mathrm{C}$` ?

a. 612 J}

b. 712 J

c. 812 J

d. 912 J

Problem 13:

How much heat is need to convert
`$0.105 \mathrm{~kg}$` of ice at
`$-10^(\circ) \mathrm{C}$` to steam at
`$140^(\circ) \mathrm{C}$` ? Use your answers from Problems 8-12.

a. 46414.5 J

b. 56414.5 J

c. 66414.5 J

d. 76414.5 J