Question

We know that the specific heat of water c = 1 calorie/g/oC For water, the latent heat of evaporation is 540 calories per gram, and latent heat of melting (or freezing) is 80 calories per gram. Answer the following questions Question 1: How much heat would be required to heat 1 gram of pure liquid water from 10oC to 20oC?

Answer 1

10 calories of heat would be required to heat 1 gram of pure liquid water from 10°C to 20°C.

Step-by-step explanation:

To calculate the heat required to heat 1 gram of pure liquid water from 10°C to 20°C, we need to use the specific heat formula:

Q = m * c * ΔT

where Q is the heat required, m is the mass of the substance, c is the specific heat of the substance, and ΔT is the change in temperature.

In this case, m = 1 gram, c = 1 calorie/g/°C, and ΔT = (20°C - 10°C) = 10°C. Substituting these values into the formula, we get:

Q = 1 gram * 1 calorie/g/°C * 10°C

Q = 10 calories

Therefore, 10 calories of heat would be required to heat 1 gram of pure liquid water from 10°C to 20°C.