Question

■ what quantity of heat is required to raise the temperature of 50.00 ml of water from 25.52 °c to 28.75 °c? the density of water at this temperature is 0.997 g/ml.

Answer 1

To find the quantity of heat needed to raise the temperature of 50.00 mL of water from 25.52 °C to 28.75 °C, one would calculate using the formula q = mcΔT, resulting in a total of 669.21 J.

Step-by-step explanation:

To calculate the quantity of heat required to raise the temperature of 50.00 mL of water from 25.52 °C to 28.75 °C, we must use the formula q = mcΔT, where:

• q is the quantity of heat (in joules)
• m is the mass of the water (in grams)
• c is the specific heat capacity of water (4.184 J/g°C)
• ΔT is the change in temperature (in °C)

First, we convert the volume of water to mass using the density given: 50.00 mL × 0.997 g/mL = 49.85 g.

Next, we calculate the change in temperature: 28.75 °C - 25.52 °C = 3.23 °C.

Finally, we plug these values into the formula:

q = (49.85 g) × (4.184 J/g°C) × (3.23 °C)

q = 669.21 J

Therefore, the quantity of heat required is 669.21 J (joules).