A 350 gram aluminum part at an initial temperature of 25.0 ° C absorbs 6200 J of heat. What is the final temperature of the aluminum part?
C(Al) = 0.900 J/g × ° C

Question

Quizplus · Accepted Answer

The heat absorbed (Q) is related to the mass (m), specific heat capacity (C), and change in temperature (ΔT) by the formula Q = m * C * ΔT. Solving for ΔT gives ΔT = Q / (m * C). Substituting the given values: ΔT = 6200 J / (350 g * 0.900 J/g°C) = 19.7 °C. Adding this to the initial temperature: 25.0 °C + 19.7 °C = 44.7 °C.

A 350 Gram Aluminum Part at an Initial Temperature of 25.0