Using the Formula Q = mCΔt

Substance | Specific Heat
(J/g°C) |
---|---|

H_{2}O (l) |
4.184 |

H_{2}O (steam) |
2.02 |

Al (s) | 0.89 |

Fe (s) | 0.45 |

How much heat is required to raise the temperature of 53 g of water from 11°C to 44°C?

To solve for heat when the temperature changes, use the equation: Q = m x C x Δt

where Q= heat which is measured in Joules (J)

m = mass which is measured in grams (g)

C = specific heat which is measured in Joules / gram °Celsius (J / g °C)

Δt = change in temperature t_{f} – t_{i} measured in degrees Celsius (°C)

**Step 1:** Identify what is given in the problem.

m = 53 g

If the substance is known, the value of C can be found on a chart like the one above.

C = 4.184 J/g°C

The change in temperature is found by subtracting the initial temperature from the final temperature:

Δt = t_{f} – t_{i} = 44°C-11°C

Δt = 33°C

**Step 2:** Plug in these values into the heat equation.

Q = m x C x Δt

Q = 53 g x 4.184 J/g°C x 33°C

Q = 7300 J

How much heat is released when 21 g of Al cools from 31.0°C to 27.0°C?

**Step 1:** Identify what is given in the problem.

m = 21 g

If the substance is known, the value of C can be found on a chart like the one above.

C = 0.89 J/g°C

The change in temperature is found by subtracting the initial temperature from the final temperature:

Δt = t_{f} – t_{i} = 27.0°C – 31.0°C

Δt = -4.0°C

Did you get a negative number for Δt? If so, you did it right! Δt = 27°C -31°C = -4.0°C. This negative number just means that heat was released.

**Step 2:** Plug in these values into the heat equation.

Q = m x C x Δt

Q = 21 g x 0.89 J/g°C x -4.0°C

Q = 75 J released (or – 75 J)

What is the specific heat capacity of 32.3 g of a metal if 50. J of heat increases the temperature of the metal by 3.5°C?

**Step 1:** Identify what is given in the problem.

Q = 50. J

m = 32.3 g

The change in temperature is given here.

Δt = 3.5°C

**Step 2:** Plug in these values into the heat equation and solve for the unknown value.

Q = m x C x Δt

C = Q/(m x Δt)

C = 50 J/ (32.3 g x 3.5°C)

C = 0.44 J/g°C

How much heat is required to change 25.0 g of liquid water into steam (water vapor)?

Changing from liquid to gas is a phase change. There is NO change in temperature during phase changes, so the equation we have been using cannot be used here. Recall, if you change from liquid to gas, the equation is Q = m x ΔHvap. Changing from a solid to a liquid uses the equation, Q = m x ΔHfus. Both ΔHvap and ΔHfus can be found on a chart. Be careful to choose the correct one!

Water |
---|

ΔHfus = 334 J/g |

ΔHvap = 2260 J/g |

In this problem, water is being changed from liquid to steam. This will require the heat of vaporization to be used.

Q = m x ΔH_{vap}

Q = 25.0 g x 2260 J/g

Q = 56,500 J

How much heat is required to change 25.0 g of ice into liquid water?

Ice is solid, so the equation to use is Q = m x ΔH_{fus}

Q = 25.0 g x 334 J/g

Q = 8350 J

How much heat is released when 155 g of water at 0°C changes to ice at the same temperature?

You already know that the quantity of heat needed to change a unit mass of solid to liquid at a constant temperature is called the heat of fusion at that temperature. For a substance, exactly the same amount of heat is released when a unit mass of liquid is changed to a solid at a constant temperature. This heat is called the heat of crystallization. FOR ANY GIVEN SUBSTANCE, THE HEAT OF CRYSTALLIZATION IS NUMERICALLY EQUAL TO THE HEAT OF FUSION!

Water |
---|

ΔH_{fus} = 334 J/g |

ΔH_{crystallization} = 334 J/g |

Q = m x ΔH_{crystallization}

Q = 155 g x 334 J/g

Q = 51800 J

Calculate the heat that is evolved when 211 g of steam at 100°C condenses to form water at the same temperature.

You already know that the quantity of heat needed to change a unit mass of liquid to gas at a constant temperature is called the heat of vaporization at that temperature. For a substance, exactly the same amount of heat is released when a unit mass of vapor is changed to liquid at a constant temperature. This heat is called the heat of condensation. FOR ANY GIVEN SUBSTANCE, THE HEAT OF CONDENSATION IS NUMERICALLY EQUAL TO THE HEAT OF VAPORIZATION!

Water |
---|

ΔH_{vap} = 2260 J/g |

ΔH_{condensation} = 2260 J/g |

Q = m x ΔH_{condensation}

Q = 211 g x 2260 J/g

Q = 477000J