Question

if 125 cal of heat is applied to 60.0 g piece of copper at 22.0 c, what will the final temperature be? The specific heat of cooper is 0.0920 cal/g

Answer 1

Heat gained or removed in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1)
125 = 60(0.0920)(T2 - 22)
T2 = 44.64 degrees Celsius

Answer 2

Answer:

The final temperature will be 44.64 °C

Explanation:

Step 1: Data given

Amount of heat = 125 cal

Mass of copper = 60.0 grams

Initial temperature of copper = 22.0 °C

Specific heat of copper = 0.0920 cal/g

Step 2: Calculate the final temperature

Q = m*c*ΔT

⇒ Q = the heat transfer = 125 cal

⇒ m = the mass of copper = 60.0 grams

⇒ c = the specific heat of copper = 0.0920 cal/g°C

⇒ ΔT = the change in temperature = T2 - T1 = T2 - 22.0 °C

125 cal = 60.0 g * 0.0920 cal/g *  ΔT

ΔT = 22.64

ΔT = T2 - T1

T2 = 22.64 + 22.0 °C

T2 = 44.64 °C

The final temperature will be 44.64 °C