The table shows the mass and melting point of some substances.

Question

The table shows the mass and melting point of some substances.

Melting Point of Substances
Mass
(kg) Melting Point of Tin
(°C) Melting Point of Lead
(°C)
1 W Y
2 232 327

Part 1: What will be the values of W and Y for 1 kg of each substance?

Part 2: Explain your answer for Part 1.

Answer 1

Part 1: The values of W and Y for 1 kg of each substance are as follows:

For Tin:
W = 1 kg
Y = 232 °C

For Lead:
W = 1 kg
Y = 327 °C

Part 2: The values for W and Y are the same for each substance regardless of their mass. This is because the question asks for the values for 1 kg of each substance specifically. The mass of the substance does not change the values of W and Y.

Answer 2

Part 1: To determine the values of W and Y for 1 kg of each substance, we need to look at the given information in the table.

From the table, we can see that for 2 kg of substance, the melting point of tin is given as 232°C and the melting point of lead is given as 327°C.

Assuming there is a linear relationship between mass and melting point, we can calculate the values of W and Y for 1 kg of each substance.

For tin:
Given mass = 2 kg, melting point = 232°C
We can set up a proportion:
2 kg / 232°C = 1 kg / Y
Cross-multiplying, we get:
2 kg × Y = 232°C × 1 kg
2Y = 232
Dividing both sides by 2, we find:
Y = 232 / 2 = 116°C

So, for 1 kg of tin, the melting point (Y) is 116°C.

For lead:
Given mass = 2 kg, melting point = 327°C
Using the same reasoning, we can set up a proportion:
2 kg / 327°C = 1 kg / W
Cross-multiplying, we get:
2 kg × W = 327°C × 1 kg
2W = 327
Dividing both sides by 2, we find:
W = 327 / 2 = 163.5°C

So, for 1 kg of lead, the melting point (W) is 163.5°C.

Part 2: The values of W and Y can be calculated using the principle of proportionality. We assume that there is a linear relationship between the mass and the melting point of substances. By setting up proportions, we can find the values for W and Y for 1 kg of each substance based on the given data for 2 kg.

Answer 3

Part 1: To find the values of W and Y, we need to analyze the given table. From the table, we are given the mass and melting point of two substances, Tin and Lead. We are asked to determine the values of W and Y for 1 kg of each substance.

Based on the table, we can see that when the mass of Tin is 2 kg, the melting point is 232°C. Since the table doesn't provide information about the mass and melting point of Tin for 1 kg, we need to make a proportional comparison to find the value of W.

We can set up a proportion using the given information:

Mass of Tin / Melting Point of Tin = 2 kg / 232°C

Now, we can use this proportion to find the value of W for 1 kg of Tin:

1 kg / W = 2 kg / 232°C

Next, we can cross multiply:

1 kg * 232°C = 2 kg * W

232 = 2W

W = 232 / 2

W = 116

So, for 1 kg of Tin, the value of W is 116.

Now, let's move on to finding the value of Y for 1 kg of Lead. From the table, we see that when the mass of Lead is 2 kg, the melting point is 327°C. Again, we can set up a proportion using this information:

Mass of Lead / Melting Point of Lead = 2 kg / 327°C

Now, we can use this proportion to find the value of Y for 1 kg of Lead:

1 kg / Y = 2 kg / 327°C

Cross multiplying:

1 kg * 327°C = 2 kg * Y

327 = 2Y

Y = 327 / 2

Y = 163.5

So, for 1 kg of Lead, the value of Y is 163.5°C.

Therefore, the values of W and Y for 1 kg of each substance are:

W = 116
Y = 163.5

Part 2: The values of W and Y in Part 1 were determined using proportions. Proportions compare two ratios or fractions and state that they are equal. By setting up proportions with the given information from the table, we were able to find the missing values of W and Y for 1 kg of Tin and Lead.