Answer to Question #102141 in Mechanical Engineering for Usman

Question

A student is sitting in the sun reading the exam. He has brought with him a bottle of 1.00 L of water. After 2 hours the temperature of the water has increased from 4 ° C to 20 ° C. During these two hours the sun / surroundings have supplied the water -66.90 kJ. Assume that the volume of 1.00 kg of water is 1.00 L and calculate the heat capacity of the water.

Answer 1

According to the heat equation:

Q = m × c × ΔT = m × c × (T₂ - T₁)

where Q - heat, m - mass, c - heat capacity, T₂ - final temperature, T₁ - initial temperature.

From here:

c = Q / [m × (T₂ - T₁)]

As a result, the heat capacity of the water is:

c = 66.90 kJ / [1.00 kg × (20 °C - 4 °C)] = 4.18 kJ/kg °C = 4.18 J/g °C

Answer: 4.18 J/g