Determine how much energy it will take to impart 200g ice taken at a temperature of -10

Determine how much energy it will take to impart 200g ice taken at a temperature of -10 in order to melt it and heat the water to 40 degrees

m = 200 g = 0.2 kg.

t1 = -10 ° C.

Cl = 2100 J / kg * ° C.

q = 3.4 * 10 ^ 5 J / kg.

t2 = 0 ° C.

t3 = 40 ° C.

Cw = 4200 J / kg * ° C.

Q -?

The ice must first be heated to the melting temperature, which is t2 = 0 ° C, then at this temperature it must be transferred to water, and the resulting water must be heated.

The required amount of heat will be the sum: Q = Q1 + Q2 + Q3.

Q1 = Cl * m * (t2 – t1).

Q1 = 2100 J / kg * ° C * 0.2 kg * (0 ° C – (-10 ° C)) = 4200 J.

Q2 = q * m.

Q2 = 3.4 * 10 ^ 5 J / kg * 0.2 kg = 68000 J.

Q3 = Cв * m * (t3 – t2).

Q3 = 4200 J / kg * ° C * 0.2 kg * (40 ° C – 0 ° C) = 33600 J.

Q = 4200 J + 68000 J + 33600 J = 105800 J.

Answer: to obtain water of the required temperature, it is necessary to tell the ice Q = 105800 J of thermal energy.