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Question 1 : The temperature of a 0.555 kg block of ice is lowered to -120°C....

Question 1 : The temperature of a 0.555 kg block of ice is lowered to -120°C. Heat (0.5159 MJ) is then transferred to the block of ice (assume ice is otherwise thermally isolated). What is the final temperature of the water in °C?

Question 2 : Calculate the time needed (seconds) to that up 285 mL of tomato soup from 4.0°C to 57°C in a 1100. watt microwave oven. Assume that the density of tomato soup is 1.25 g/mL and that its specific heat capacity is 4.21 J/g°C.

Question 3 : The tallest building on Earth in 2016 is the Burj Khalifa in Dubai, which stands at 2717 ft tall. A 15.0 kg iron sphere is dropped from the very top of the Burj Khalifa building. What is the theoretical maximum temperature increase possible in degrees Celsius for the iron sphere if measured a minute distance above the ground before impact?

Question 4 : You suspect that your water supply is contaminated with bacteria so decide to boil 15.0 gallons of water. A sudden power outage in your home forces you to use camping gas (propane) as a fuel to boil the water. What volume of fuel (litres) will you need to boil the water if its initial temperature is 22.5°C? (the density of liquid propane is 0.493 g/cm

^3​3​​. Only 17% of the heat of combustion of propane goes towards heating the water).

Question 5 : The Dulong-Petit law states that the specific heat capacity (c) of any element is equal to 3R/M, where R is a constant equal to 8.314 J/mol K, and M is the molar mass of the element measure in units of g/mol. Use specific heat capacity data from table 5.2 to plot the Dulong-Petit law as a straight line graph. Use the equation from your straight line graph to identify the element whose specific heat is 0.137 J/g°C.

I really need help!!!

Question 1 : The temperature of a 0.555 kg block of ice is lowered to -120°C. Heat (0.5159 MJ) is then transferred to the block of ice (assume ice is otherwise thermally isolated). What is the final temperature of the water in °C?

We require 2 type of heat, latent heat and sensible heat

Sensible heat (CP): heat change due to Temperature difference

Latent heat (LH): Heat involved in changing phases (no change of T)

Then

Q1 = m*Cp ice * (Tf – T1)

Q2 = m*LH ice

Q3 = m*Cp wáter * (Tfinal– Tfusion)

Note that Tf = 0°C; Tb = 100°C, LH ice = 334 kJ/kg;

Cp ice = 2.01 J/g°C ; Cp water = 4.184 J/g°C

Then

Q1 = 555*2.01 * (0 – -120)

Q2 = 555*334

Q3 = 555*4.184 * * (Tfinal– 0)

0.5159*10^6 = 555*(2.01*(120) + (334) + 4.184*Tf)

(0.5159*10^6) - 555*(2.01*(120) + (334)) = 555*4.184*Tf

196664 = 555*4.184*Tf

Tf = 196664 /(555*4.184) = 84.69°C

Tfinal of block --> melted = 84.69°C