Answer:
It's just asking you to rearrange to equations to get what you want, so no numbers involved it's just teaching you the basics.
Explanation:
Answer:
T2 = 29°C
Explanation:
Given data:
Heat added = 420 j
Mass of water = 25 g
Initial temperature = 25°C
Final temperature = ?
Solution;
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water = 4.18 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
Now we will put the values.
420 j = 25 g ×4.18 j/g.°C × (Final temperature - initial temperature)
420 j = 25 g ×4.18 j/g.°C × (T2 - 25°C)
420 j = 104.5 j/°C × (T2 - 25°C)
420 j /104.5 j/°C = T2 - 25°C
4°C + 25°C = T2
T2 = 29°C
Answer:The product and reactants reach a final, unchanging level.
D 4 grams
i need to write a couple more character for explanation so there
