Need answer ASAP If 20.0g of water at 56.2°C absorbs 195 J of heat, what is the final temperature?
^in degrees celsius

What mass of water would increase in temperature from 45.0°C to 98.0°C after
absorbing 6830 J of energy?
^IN GRAMS

1. 58.5°C = Final temperature

2. Mass of water = 30.8g

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 =

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

c, liquids evaporate, creating a gas