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 =
`(-23)/(1.1)`
T ≈ - 20.90 ° C (the final temperature of the ice)