(1)calculate the heat energy datas temperatures 100degree celciuse and 4degree celciuse,mass 0.5kg take latent heat of vapourisation 2.02times10raised to the power of minus6.(2)calculate the specific latent heat of fusion mass of calorimeter 0.01kg and mass of ice 0.09kg and 0.10kg giving specific heat capacity of water 4200. Pls work it know for me am in class.

Its physic question am very sorry all.

Sure, I can help you with that! Let's break down each question and calculate the answers step by step.

(1) To calculate the heat energy, we can use the formula:

Q = mL

Where,
Q = Heat energy (in Joules)
m = Mass (in kilograms)
L = Latent heat of vaporization (in J/kg)

Given:
Temperature of water (initial) = 100 degrees Celsius
Temperature of water (final) = 4 degrees Celsius
Mass of water = 0.5 kg
Latent heat of vaporization = 2.02 x 10^(-6) J/kg

Step 1: Calculate the heat energy required to cool the water from 100°C to 4°C.
We can use the formula:

Q1 = mcΔT

Where,
m = Mass (in kilograms)
c = Specific heat capacity of water (in J/(kg·°C))
ΔT = Change in temperature (in °C)

The specific heat capacity of water is approximately 4200 J/(kg·°C).

Calculating Q1:
Q1 = (0.5 kg) × (4200 J/(kg·°C)) × (100°C - 4°C)

Q1 = 0.5 kg × 4200 J/(kg·°C) × 96°C

Q1 = 201,600 J

Step 2: Calculate the heat energy required to vaporize the water.
We can use the formula:

Q2 = mL

Calculating Q2:
Q2 = (0.5 kg) × (2.02 × 10^(-6) J/kg)

Q2 = 1.01 × 10^(-6) J

Step 3: Calculate the total heat energy required.
To find the total heat energy, we need to sum up Q1 and Q2:

Total heat energy (Q) = Q1 + Q2

Calculating Q:
Q = 201,600 J + 1.01 × 10^(-6) J

Q = 201,601 J

Therefore, the heat energy required to go from 100°C to 4°C and then vaporize 0.5 kg of water is 201,601 Joules.

(2) To calculate the specific latent heat of fusion, we can use the formula:

L_fusion = (m_c + m_w) × c_w × ΔT / m_ice

Where,
L_fusion = Specific latent heat of fusion (in J/kg)
m_c = Mass of calorimeter (in kilograms)
m_w = Mass of water (in kilograms)
c_w = Specific heat capacity of water (in J/(kg·°C))
ΔT = Change in temperature (in °C)
m_ice = Mass of ice (in kilograms)

Given:
Mass of calorimeter (m_c) = 0.01 kg
Mass of ice (m_ice) = 0.09 kg
Mass of water (m_w) = 0.10 kg
Specific heat capacity of water (c_w) = 4200 J/(kg·°C)

Step 1: Calculate the change in temperature of the water.
The ice will melt, and the water and ice will reach thermal equilibrium. The final temperature will be 0°C. So, ΔT = 0°C - (-10°C) = 10°C.

Step 2: Calculate the specific latent heat of fusion.
Using the given formula:

L_fusion = (0.01 kg + 0.10 kg) × 4200 J/(kg·°C) × 10°C / 0.09 kg

L_fusion = 5000 J/kg

Therefore, the specific latent heat of fusion is 5000 J/kg.

I hope this helps! Let me know if you have any further questions.