Answer:
Q = 30355.2 J
Explanation:
Given data:
Mass of ice = 120 g
Initial temperature = -5°C
Final temperature = 115°C
Energy required = ?
Solution:
Specific heat capacity of ice is = 2.108 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Q = m.c. ΔT
ΔT = T2 -T1
ΔT = 115 - (-5°C)
ΔT = 120 °C
Q = 120 g × 2.108 j/g.°C × 120 °C
Q = 30355.2 J
Answer:
904.014 j/kgk
Explanation:
Mass of metal = 45g
Temperature of metal = 85.6°
Mass of water = 150
Temperature of water = 24.6
Final temperature of system = 28.3
Heat lost by metal = Heat gained by water
m1 * c1 * dt = m2 * c2 * dt
Q = quantity of heat
Q = m*c*dt
dt = change in temperature
dt of water = 28.3 - 24.6 = 3.7
dt of metal = 85.6 - 28.3 = 57.3
Specific heat capacity of water, c = 4200
(45 * 10^-3) * c * 57.3 = (150 * 10^-3) * 4200 * 3.7
2.5785c1 = 2331
c1 = 2331 / 2.5785
= 904.01396
= 904.014 j/kgk
Answer:
Sun spots are cooler spots on the sun so anything to cool off part of the sun would create a sunspot
Hope this helps and have a great day!
Explanation:
