Question

Воду массой 800 грамм при температуре 25 градусов смешали с кипятком массой 200 грамм. Температура полученной смеси стала равна 40 градусам. Найти количество теплоты, которое участвовало в теплообмене.

Answer 1

Q1=mC(T2-T1)=0,8*4200*(40-25)= 50400 Дж
Q2=Q1=50400 Дж
Q1+Q2=100,8 кДж - это ответ

Answer 2

Calculating Heat Exchange in Mixing Water

To calculate the amount of heat involved in the exchange, we can use the formula:

Q = mcΔT

Where: - Q = heat energy - m = mass of the substance - c = specific heat capacity - ΔT = change in temperature

Given: - Mass of water (m1) = 800 grams - Initial temperature of water (T1) = 25 degrees Celsius - Mass of boiling water (m2) = 200 grams - Final temperature of the mixture (Tf) = 40 degrees Celsius

First, we need to find the specific heat capacity of water, which is approximately 4.18 J/g°C.

Using the formula Q = mcΔT for both the water and boiling water, we can calculate the heat exchange for each and then sum them to find the total heat exchange.

For water: Q1 = m1cΔT1

For boiling water: Q2 = m2cΔT2

Where ΔT1 = Tf - T1 and ΔT2 = Tf - 100 (since boiling water is at 100 degrees Celsius).

Let's calculate the heat exchange for each and then sum them to find the total heat exchange.