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

500cal

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

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