Question

What is the specific heat of material if 510 calories is required to raise the temperature of 170 grams of a material from 45°C to 75°C​

Answer 1

Given - Heat energy - 510 calories

Mass of material - 170 gram

Initial temperature - 45° C

Final temperature - 75° C

Find - Specific heat of material.

Solution - The formula for specific heat is represented as = Q = m*c*∆t. Q stands for heat energy in Joule, m represents mass of material in kilogram, c is specific heat and ∆t represents change in temperature in Kelvin.

1 calorie is equivalent to 4.184 joules

510 calories = 2133.8 joules.

Heat energy = 2133.8 joules.

Mass of material = 170 grams = 0.17 kg

Initial temperature = 45° C = 318 K

Final temperature = 75° C = 348 K.

Keeping the values in equation-

C = Q/m(t2-t1)

C = 2133.8/0.17*(348-318)

C = 2133.8/0.17*30

C = 2133.8/5.1

C = 418.4 J/kg K

Hence, the specific heat of material is 418.4 J/kg K.