Answer:
2274 J/kg ∙ K
Explanation:
The complete statement of the question is :
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
= mass of metal = 400 g
= specific heat of metal = ?
= initial temperature of metal = 100 °C
= mass of aluminum cup = 100 g
= specific heat of aluminum cup = 900.0 J/kg ∙ K
= initial temperature of aluminum cup = 15 °C
= mass of water = 500 g
= specific heat of water = 4186 J/kg ∙ K
= initial temperature of water = 15 °C
= Final equilibrium temperature = 40 °C
Using conservation of energy
heat lost by metal = heat gained by aluminum cup + heat gained by water
Answer:
Antarctic Circle
Explanation:
The Tropic of Cancer, which is also referred to as the Northern Tropic, is the most northerly circle of latitude on Earth at which the Sun can be directly overhead. This occurs on the June solstice, when the Northern Hemisphere is tilted toward the Sun to its maximum extent.
Tropic of Capricorn Is it Southern Hemisphere counterpart, marking the most southerly position at which the Sun can be directly overhead.
D. 289
Take the formula:
K=5/9(Fahrenheit-32)+273
Plug in Fahrenheit
K=5/9 (60-32)+273
From here it is simple math and you can plug it into your calculator getting 288.5555556 and round to 289
