Question

An ice cube of mass 0.1 kg at 0^@C is placed in an isolated container which is at 227^@C. The specific heat s of the container varies with temperature T according to the empirical relation s=A+BT, where A= 100 cal//kg-K and B = 2xx (10^-2) cal//kg-K^2. If the final temperature of the container is 27^@C, determine the mass of the container. (Latent heat of fusion for water = 8xx (10^4) cal//kg, specific heat of water =10^3 cal//kg-K).

Answer 1

Answer:

In Euclidean geometry, a quadrilateral is a four-sided 2D figure whose sum of internal angles is 360°. The word quadrilateral is derived from two Latin words ‘quadri’ and ‘latus’ meaning four and side respectively. Therefore, identifying the properties of quadrilaterals is important when trying to distinguish them from other polygons. So, what are the properties of quadrilaterals? There are two properties of quadrilaterals:

A quadrilateral should be closed shape with 4 sides

All the internal angles of a quadrilateral sum up to 360°

In this article, you will get an idea about the 5 types of quadrilaterals and get to know about the properties of quadrilaterals.

This is what you’ll read in the article:

Different types of quadrilaterals

Rectangle

Properties of rectangles

Rectangle formulas

Square

Properties of a square

Formulas of squares

Parallelogram

Properties of parallelograms

Parallelogram formulas

Rhombus

Properties of a rhombus

Rhombus formulas

Answer 2

The mass of the container is 0.495 kg .

Explanation:

According to calorimetry principle,

Heat lost by container = Heat gained by ice

T = 273 + 27 = 300 K

Heat lost by container:

Let dQ be the heat lost when the temperature of the container is 'T'.

dQ = mcdT

Where,

m = Mass of the container

c = A + BT

Now, the heat lost is given by the formula:

dQ = m(A + BT)dT

On integrating, we get,

Q = $\int_{500}^{300}$ m(A + BT)dT

Q $=m\left[A T+\frac{(B T)^{2}}{2}\right]_{500}^{300}$

Heat lost = Q = - 21600 × m Calorie

Heat gained by the ice:

0° ice → 0° water

0° water → 27° water

Q₁ = mLQ₂ = mc∆T

0.1 × 80,000 = 0.1 × 10³ × 27

8000 cal = 2700 cal

∴ Q₁ + Q₂ = 8000 + 2700 = 10,700 cal

Heat lost = heat gained

21600 m = 10,700

∴ m = 0.495 kg