A 35.0 g piece of metal wire is heated, and the temperature changes from 21 degrees Celsius to 52 degrees Celsius. The amount of heat absorbed is 320 calories. What is the specific heat of the metal?

1.23 j/g. °C

Explanation:

Given dа ta:

Mass of metal = 35.0 g

Initial temperature = 21 °C

Final temperature = 52°C

Amount of heat absorbed = 320 cal (320 * 4.184 = 1338.88 j)

Specific heat capacity of metal = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 52°C - 21 °C

ΔT = 31°C

1338.88 j = 35 g * c * 31°C

1338.88 j = 1085 g.°C * c

1338.88 j/1085 g.°C = c

1.23 j/g. °C = c