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

d.7.5cm is your answer

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach