If 10 calories of energy are added to 2 grams of ice at -30° C, calculate the final temperature of the ice. (Notice that the specific heat of ice is different from that of water.)

Verified Answer :


At where:

 
Q = amount of sensible heat (cal or J).

 
C = specific heat of the substance constituting the body (cal / g ° C or J / kg ° C).

 
M = body mass (g or kg).

 
Δθ = temperature variation (° C).

 
T = final temperature

 
To = Initial temperature

 
Data:


Q = 10 calories

M = 2 grams

C (ice)= 0.550 cal / g ° C

To = -30 ° C

T =?

 
Formula:

Q = m c Δθ

 
Resolution:

Substitute
 Q = m c Δθ

10 = 2 0.550 [T-To] 
10 = 1.1 [T-(-30)] 
10 = 1.1
[T+30]
10 = 1.1T + 33
-1.1T = 33 - 10
-1.1T = 23 .(-1)
1.1T = - 23 
T = \(\frac{-23}{1.1}\) 
T ≈ - 20.90 ° C (the final temperature of the ice)