Determine how much heat is needed to transform 200 g of ice at a temperature of -10 ° C
Determine how much heat is needed to transform 200 g of ice at a temperature of -10 ° C into water with a temperature of 20 ° C.
Given:
m = 200 grams = 0.2 kilograms is the mass of ice;
T0 = -10 degrees Celsius – ice temperature;
T1 = 0 degrees Celsius – ice melting temperature;
T2 = 20 degrees Celsius – required water temperature;
c1 = 2050 J / (kg * C) – specific heat capacity of ice;
q = 3.4 * 10 ^ 5 J / kg – specific heat of ice melting;
c2 = 4200 J / (kg * C) – specific heat capacity of water.
It is necessary to calculate the amount of heat Q (Joule) that will be required to melt the ice and heat the water to the temperature T2.
We find the total amount of heat by the formula:
Q = Q ice heating + Q ice melting + Q water heating.
Q = c1 * m * (T1 – T0) + q * m + c2 * m * (T2 – T) =
= 2050 * 0.2 * (0 + 10) + 3.4 * 10 ^ 5 * 0.2 + 4200 * 0.2 * (20 – 0) =
= 410 * 10 + 6.8 * 10 ^ 4 + 840 * 20 =
= 4100 + 68000 + 16800 = 88900 Joules.
Answer: it is necessary to spend an amount of heat equal to 88,900 Joules.
