*Q = M C ΔT* formula was conceived to calculate quantities of sensible heat Q. But it can be manipulated to calculate heated masses or differences in temperature, once the other values are known:

*These
are not new formulas; simply variations of the formula Q = M C ΔT,
depending onwhat we know, and what we are looking for.*

The units used must obviously be the same whatever the presentation of the formula, thus:

*Q in [kJ] **M: mass in [kg]**C in [kJ/kg.K] **ΔT in [K]*

#### Question

What mass of water in [kg] could be heated by 40 [K] if 162,000 [kJ] (153,546 Btu) were available?

968.9 [kg]

Explanation:

The specific heat of water is 4.18 [kJ/kg.K]

The mass of water that could be heated by de 40 [K], with 162,000 [kJ] will be:

162,000/(4.18 x 40) = 968.9 [kg]

#### Question

By how many degrees in [K] could we heat 3,000 [l] of water with a supply of 540,000 [kJ]?

43 [K]

Explanation:

The specific heat of water is 4.18 [kJ/kg.K] (3.96 Btu/kg.K)

The considered density of water is 1,000 [kg/m^{3}].

Hence 3,000 [l] = 3,000 [kg]

With a supply of 540,000 [kJ] the rise in temperature of 3,000 [l] will be:

540,000/(3,000 x 4.18) = 43 [K]

#### Question

With the specific heat of air being 1 [kJ/kg.K], what mass of air (in [kg]) could we heat from 10 to 35 [°C] with a supply of 15,000 [kJ] (14,217 Btu)?

600 [kg]

500 [m^{3}]Explanation:

The mass of air heated from 10 to 35 [°C] with a supply of 15,000 [kJ] will be:

15,000/(1 x 25) = 600 [kg]

Hence the volume of air heated will be 600/1.2 = 500 [m^{3}]

#### Question

The specific heat of domestic fuel is 2.1 [kJ/kg.K] (1.99 Btu/kg.K).

The density of domestic fuel is 840 [kg/m^{3}].

What quantity of heat needs to be applied to a tank of 25 [m^{3}] to heat it from 5 to 15 [°C]?

441,000 [kJ] (417,987 Btu)

Explanation:

The mass of domestic fuel that can be heated is: 25 x 840 = 21,000 [kg]

The quantity of heat to be applied to the tank to raise the temperature of the fuel by 10 [K] will be: 21,000 x 2.1 x 10 = 441,000 [kJ]

#### Question

The specific heat of domestic fuel is 2.1 [kJ/kg.K] (1.99 Btu/kg.K).

What mass of domestic fuel could be heated by 25 [K], with 864,000 [kJ] (818,913 Btu)?

Knowing that the density of domestic fuel is 840 [kg/m^{3}], what volume of fuel (in [m^{3}]) could be heated?

16 457 [kg]

19.59 [m^{3}]

Explanation:

The mass of domestic fuel to be heated will be: 864,000/(2.1 x 25) = 16,457 [kg]

The volume of fuel heated will be: 16,457/840 = 19.59 [m^{3}]

#### Question

The specific heat of domestic fuel is 2.1 [kJ/kg.K] (1.99 Btu/ kg K).

The density of domestic fuel is 840 [kg/m^{3}].

By how many degrees [°C] or ([K]) could we heat a tank of 6 [m^{3}], with 72,000 [kJ] (68,242 Btu)?

6.8 [K]

Explanation:

The mass of domestic fuel to be heated will be:

6 x 840 = 5,040 [kg]

The rise in temperature of this mass of domestic fuel with 72,000 [kJ] available will be:

72,000/(5,040 x 2.1) = 6.8 [K]

#### Question

By how many degrees [K] could we heat 1 [kg] of water if 1,000 Joules (0.95 Btu) are supplied?

0.24 [K]

Explanation:

The specific heat of water is 4.18 [kJ/kg.K] (3.96 Btu/kg.K)

1,000 Joules corresponds to 1 [kJ]

The rise in temperature of 1 [kg] of water with a supply of 1 [kJ] will be:

1/4.18 = 0.24 [K]