Question

an object with a heat capacity of 3.40×103j∘c absorbs 54.0 kj of heat, beginning at −25.0∘c. what will be the final temperature of the object?

Answer 1

the final temp would be -9.12 °C

Step-by-step explanation:

use Q = mcΔT equation

Given:

Heat capacity (C) = 3.40 × 10^3 J/°C

Heat absorbed (Q) = 54.0 kJ = 54.0 × 10^3 J

Initial temperature (T1) = -25.0 °C

rearrange the equation since we are solving for temperature...

ΔT = Q / (mc)

so...

ΔT = (54.0 × 10^3 J) / (1 × 3.40 × 10^3 J/°C)

then...

ΔT ≈ 15.88 °C

to find the final temperature,just simply add the temperature to the initial temperature...

Final temp (T2) = T1 + ΔT

(T2) = -25.0 °C + 15.88 °C

so the answer would be...

(T2) ≈ -9.12 °C