Question

The specific heat of a substance which absorbs 7.5 * 10^3 J of heat when sample with mass of 1.0 *10^5 g of the substance increases in temperature from 10 degrees Celsius to 90 degrees Celsius will be:

Answer 1

The specific heat of the substance which absorbs 7.5 * 10^3 J of heat when its temperature increases from 10°C to 90°C with a mass of 1.0 * 10^5 g is 0.09375 J/g°C.

Step-by-step explanation:

The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by 1°C. To determine the specific heat in this case, we will use the formula:

Specific Heat (c) = Q / (m × ΔT)

Where:

• Q is the amount of heat absorbed, in joules (J)
• m is the mass of the substance, in grams (g)
• ΔT is the change in temperature, in degrees Celsius (°C)

Given:

• Q = 7.5 × 103 J
• m = 1.0 × 105 g
• ΔT = 90°C - 10°C = 80°C

Plugging these into the formula gives us:

c = (7.5 × 103 J) / (1.0 × 105 g × 80°C)

After calculating, we find that the specific heat of the substance is:

c = 0.09375 J/g°C

Therefore, the specific heat of the substance is 0.09375 J/g°C.