Question

A 50-gram sample of water is initially at a temperature of 22 °C. The sample is heated until the temperature is 32 °C The specific heat of water is 1.00 cal/g °C. How much heat is absorbed by the water in calories? Enter your answer without a decimal​

Answer 1

500cal

Step-by-step explanation:

Given parameters:

Mass of water = 50g

Initial temperature = 22°C

Final temperature = 32°C

Specific heat of water = 1cal/g

Unknown:

Amount of heat absorbed by the water in calories = ?

Solution:

To solve this problem, we use the expression below:

H = m c Ф

H is the amount of heat absorbed

m is the mass

c is the specific heat capacity

Ф is the temperature change

H = 50 x 1 x (32 - 22) = 500cal