Question

A pan containing 40 grams of water was allowed to cool from a temperature of 91 degrees C. If the amount of heat released is 1300 J, what is the final temp

Answer 1

Answer:

Final temperature is 83.236 °C.

Explanation:

Let final temperature to be 'T'

There is a theorem :

ΔQ = m·s·ΔT

ΔQ : Heat exchange = -1300 J (it is negative because heat is released)

m : mass which exchanges heat = 40 grams

s : specific heat capacity of water = 4.186 J/grams °C = 1 Calories/grams °C

ΔT : Change in temperature = T - 91°

-1300 = 40 × 4.186 × (T-91)

$\mathrm{(T-91)=-\frac{1300}{40\times4.186}}$

$\mathrm{(T-91)\approx-7.764}$

$\mathbf{T\approx83.236 \ ^{0}C}$