<span>1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules
Formula: Heat = mass * specific heat * ΔT
Data:
mass = 50 g = 0.050 kg
specific heat of iron = 450 J/ kg °C
ΔT = 124°C - 10°C ¿ 114 °C
=> heat = 0.050kg * 450 J / kg°C * 114°C ≈ 2.6 J
2. If a refrigerator is a heat pump that follows the first law of
thermodynamics, how much heat was removed from food inside of the
refrigerator if it released 492J of energy to the room? Joules
The firs law of thermodynamics is conservation of energy => energy removed from inside of the refrigerator = energy released to the room
=> Answer = 492 J
3. How much heat is needed to raise the temperature of 45g of water by 63oC? Joules
Formula: heat = mass * specific heat * ΔT
specific heat of water = 4186 J / Kg °C
heat = 0.045 kg * 4186 J/kg °C * 63°C = 11,867.31 J
</span>
Answer:
A. 1,156
Step-by-step explanation:
The equation can be..
x= 1,156÷34
The sum of this equation would lead to x=34, since 34×34 is 1,156.
Answer:
3f+8g
Step-by-step explanation:
<h2>For the number 1<u>.</u><u>.</u></h2>
<h2><u>angle </u><u>at </u><u>center </u><u>=</u><u> </u><u>2</u><u>×</u><u>a</u><u>n</u><u>g</u><u>l</u><u>e</u><u> </u><u>at </u><u>circumference</u></h2><h2><u>the </u><u>angle </u><u>at </u><u>center </u><u>(</u><u>o)</u><u> </u><u>is </u><u>equal</u><u> </u><u>to</u><u> </u><u>1</u><u>4</u><u>1</u></h2>
<u>therefore:</u>
<h2><u>1</u><u>4</u><u>1</u><u> </u><u>=</u><u> </u><u>2</u><u>x</u></h2><h2><u>divide</u><u> </u><u>both </u><u>sides </u><u>by </u><u>2</u></h2><h2><u>x </u><u>=</u><u> </u><u>7</u><u>0</u><u>.</u><u>5</u></h2>
<u>option</u><u> </u><u>(</u><u>A)</u><u>.</u>
<u>(</u><u> </u><u>the </u><u>number </u><u>2</u><u> </u><u>and </u><u>3</u><u> </u><u>questions</u><u> </u><u>aren't</u><u> </u><u>correct </u><u>)</u>
Answer:
The Answer is c: 4x5 + x3 + 2x2 – 3
Step-by-step explanation:
