What is the specific heat of a substance that absorbs 2500 joules of heat when a sample of 100g of the substance increases in temperature from 10.°C to 70.°

0.417 J/gºC

Explanation:

From the question given above, the following data were obtained:

Heat (Q) absorbed = 2500 J

Mass (M) of substance = 100 g

Initial temperature (T1) = 10 °C

Final temperature (T2) = 70 °C

Specific heat capacity (C) =?

Next, we shall determine the change in temperature (ΔT). This can be obtained as follow:

Initial temperature (T1) = 10 °C

Final temperature (T2) = 70 °C

Change in temperature (ΔT) =?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 70 – 10

Change in temperature (ΔT) = 60 °C

Finally, we shall determine the specific heat capacity of the substance as follow:

Heat (Q) absorbed = 2500 J

Mass (M) of substance = 100 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) =?

Q = MCΔT

2500 = 100 × C × 60

2500 = 6000 × C

Divide both side by 6000

C = 2500 / 6000

C = 0.417 J/gºC

Therefore, the specific heat capacity of substance is 0.417 J/gºC

the temperature of the gas, in k, not celsius.

6.4 cm (max) - 25 cm (min)