What is the specific heat of a substance that absorbs 2500 joules of heat when a sample of 1.200 kg of the substance increases in temperature from 10.0°C to 70.0°C?

0.035 J/g°C

Explanation:

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

Heat (Q) absorbed = 2500 J

Mass (M) = 1.2 Kg

Initial Temperature (T₁) = 10 °C

Final Temperature (T₂) = 70 °C

Specific heat capacity (C) =?

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

Initial Temperature (T₁) = 10 °C

Final Temperature (T₂) = 70 °C

Change in temperature (ΔT) =

ΔT = T₂ – T₁

ΔT = 70 – 10

ΔT = 60 °C

Thus, the change in the temperature of the substance is 60 °C

Next, we shall convert 1.2 Kg to grams (g). This can be obtained as follow:

1 Kg = 1000 g

Therefore,

1.2 Kg = 1.2 Kg × 1000 g / 1 Kg

1.2 Kg = 1200 g

Thus, 1.2 Kg is equivalent to 1200 g.

Finally, we shall determine the specific heat capacity of substance. This can be obtained as follow:

Heat (Q) absorbed = 2500 J

Mass (M) = 1200 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) =?

Q = MCΔT

2500 = 1200 × C × 60

2500 = 72000 × C

Divide both side by 72000

C = 2500 / 72000

C = 0.035 J/g°C

Therefore, the specific heat capacity of the substance is 0.035 J/g°C

a high pressure system is a dense air mass that is usually cooler and drier than the surrounding air. a low pressure system is a less dense air mass that is usually wetter and warmer than the surrounding air.

in general, areas that experience high atmospheric pressure also experience fair weather. low pressure systems can cause the formation of clouds and storms. air usually flows from areas of high pressure to areas of low pressure..