Question

1. determine the amount of heat required to increase the temperature of 47 g from 65°c to 89°c. the specific heat capacity for water is 1 cal/g°c.

(a) 450 cal
(b) 600 cal
(c) 510 cal
(d) none of the above

1. the heat of vaporization of ethyl alcohol ( l_{v} ) is about 200 cal/g. that means that 1 gram of this refrigerant upon vaporization would release

a) less than 200 calories.
b) 200 calories.
c) more than 200 calories.

1. the quantity of heat required to change the temperature of 1 kg of a substance by 1 degree is

(a) specific heat capacity
(b) heat
(c) temperature
(d) entropy
(e) none of the above

Explanation:

Question 19

Step1: Calculate temperature change

$\Delta T=T_{final}-T_{initial}=89^{\circ}C - 65^{\circ}C=24^{\circ}C$

Step2: Use heat formula

The formula for heat is $Q = mc\Delta T$, where $m = 47g$, $c=1cal/g^{\circ}C$, and $\Delta T = 24^{\circ}C$.
Substitute values: $Q=47\times1\times24$
$Q = 1128cal$

The heat of vaporization \(L_v\) is the amount of heat required to vaporize a substance. When 1 gram of ethyl alcohol vaporizes, it absorbs (not releases) \(L_v = 200cal/g\). So the answer is based on the definition of heat of vaporization.

By definition, the specific heat capacity \(c\) is the quantity of heat required to change the temperature of \(m = 1kg\) of a substance by \(\Delta T=1^{\circ}C\), using the formula \(Q=mc\Delta T\).

Answer:

d) none of the above

Question 20