Answer:
1. 58.5°C = Final temperature
2. Mass of water = 30.8g
Step-by-step explanation:
The general equation of specific heat of a substance is:
Q = C*m*ΔT
Where Q is the heat involved in Joules,
C is specific heat (Water = 4.184J/g°C),
m is mass of substance,
ΔT is change in temperature (Final temperature - Initial temperature)
For the first problem:
1. 195J = 4.184J/g°C*20.0g*ΔT
ΔT = 2.3°C = Final T - Initial T
ΔT = 2.3°C + 56.2°C =
58.5°C = Final temperature
In the seconf problem:
2. 6830J = 4.184J/g°C*m*(98.0°C-45.0°C)
6830J / 4.184J/g°C*53°C = m
30.8g = m
Mass of water = 30.8g