Question

How much heat is released when 275g of water cools from ${85.2^0}C$ to ${38.4^0}C$ .

Answer 1

This could be simply solved by applying the concepts of heat transfer. Also, we know that the temperature is changing, and mass of substance is provided. We shall use the formula given below.

Formula used: Here, we will use the number of moles formula:
$Q = m{C_p}\Delta t$
$Q$ is the heat (in Joules)
$m$ is the mass in gm
${C_p}$ specific heat capacity
$\Delta t$ is change in temperature.

Complete step by step answer
We are given that 275g of water cools down.
We have our initial and final temperatures.
The value of ${C_p}$ is $4.18{\text{J/g}}{}^{\text{0}}{\text{C}}$
On putting in the formula we get the answer.
$Q = m{C_p}\Delta t$
$Q = 275 \times 4.18 \times (38.4 - 85.2)$
On solving further get,
$Q = - 53.79{\text{kJ}}$
Thus, the amount of heat released is $53.79{\text{kJ}}$ .

Note
It should be always kept in mind that Specific heat, ratio of the quantity of heat required to raise the temperature of a body one degree to that required to raise the temperature of an equal mass of water one degree or in other words Specific heat capacity is a measure of the amount of heat energy required to change the temperature of 1 kg of a material by 1 K. Hence it is important as it will give an indication of how much energy will be required to heat or cool an object of a given mass by a given amount.
And latent heat is defined as the heat or energy that is absorbed or released during a phase change of a substance. It could either be from a gas to a liquid or liquid to solid and vice versa. Latent heat is related to a heat property called enthalpy.