Question

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.)

Answer 1

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)