Question

A block of aluminum with a mass of 140 g is cooled from 98.4°C to 62.2°C with a release of 4817 J of heat. From these data, calculate the specific heat of aluminum?

Answer 1

specific heat = 0.951 j/g·°C

Step-by-step explanation:

Heat flow equation => q = m·c·ΔT

q = heat flow = 4817 joules

m = mass in grams = 140 grams Aluminum

c = specific heat = ?

ΔT = Temperature Change in °C = 98.4°C - 62.2°C = 36.2°C

q = m·c·ΔT => c = q/m·ΔT = 4817j/(140g)(36.2°C) = 0.951 j/g·°C