Question

What is the specific heat of a 44 g piece of metal if 202 Joules of heat are required to

Answer 1

Answer:

≈ 395,8 J/(kg * °C)

Explanation:

m = 44 g = 0,044 kg

$t_{1}$ = 22 °C

$t_{2}$ = 33,6 °C

Q = 202 J

The formula is: Q = c * m * ($t_{2} - t_{1}$)

c = $\frac{Q}{m * (t_{2} - t_{1} )}$

Calculating:

c = 202 J / 0,044 kg * (33,6 °C - 22 °C) ≈ 395,8 J/(kg * °C)