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

the components of a vector are the parts of the vectors that are to each other.

b perpendicular

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation