A vessel containing 100 g of water at a temperature of 0 ° C was brought into the room.

A vessel containing 100 g of water at a temperature of 0 ° C was brought into the room. After 15 minutes the temperature rose to 2 ° C. When there was 100 g of ice in the vessel at the same temperature, it melted in 10 hours. Determine the specific heat of melting. The heat capacity of water is considered known.

Given:

m1 = m2 = 100 grams = 0.1 kilograms – the mass of water and ice;

c = 4200 J / (kg * C) – specific heat capacity of water;

T0 = ​​0 degrees Celsius – initial water temperature;

T1 = 2 degrees Celsius – water temperature after a time interval t1;

t1 = 15 minutes – the time during which the water temperature changed from T0 to T1;

t2 = 10 hours = 600 minutes – the time it took for the ice to melt.

It is required to determine q (J / kg) – specific heat of melting of ice.

Let’s find the amount of heat that was required to heat the water:

Q = c * m1 * (T1 – T0) = 4200 * 0.1 * (2 – 0) = 420 * 2 = 840 Joules.

This energy was released during time t. That is, energy is released in the room at a rate:

v = Q / t1 = 840/15 = 56 J / minute.

Then, it took energy to melt the ice:

Q2 = v * t2 = 56 * 600 = 33600 Joules.

The specific heat of melting of ice will be equal to:

q = Q2 / m2 = 33600 / 0.1 = 336000 Joules = 336 kJ.

Answer: the specific heat of melting of ice is 336 kJ.