Question

A 52-gram sample of water that has an initial temperature of 10.0 °C absorbs 4,130 joules. If the specific heat of water is 4.184 J/(g °C), what is the final temperature of the water? (4 points) Group of answer choices 11 °C 19 °C 29 °C 51 °C

Answer 1

28.98°C

Step-by-step explanation:

Mass = 52g

Initial temperature (T1) = 10°C

Final temperature (T2) = ?

Heat energy (Q) = 4130J

Specific heat capacity of water = 4.184J/g°C

Heat energy (Q) = mc∇T

Q = heat energy

M = mass of the substance

C = specific heat capacity of the substance

∇T = change in temperature (T2 - T1)

Q = mc∇T

4130 = 52 * 4.184 * (T2 - 10)

4130 = 217.568 * ( T2 - 10)

4130 = 217.568T2 - 2175.68

217.568T2 = 4130 + 2175.68

217.57T2 = 6305.68

T2 = 6305.68 / 217.57

T2 = 28.98°C

Final temperature is 28.98°C