The temperature of 100 g of liquid water in a calorimeter changes from 25°c to 50°c. how much heat was transferred? use the equation q= mcδt. the specific heat of liquid water is 4.18 j/g-°c. 10.45kj 90.2 kj 20.9 kj 2500 kj

Answer : 10.45 kJ of heat was transferred.

Explanation :

The equation we need to use here is given as,

Where q is the amount of heat transferred

m is the mass of the water

C is the specific heat of water

ΔT is the change in temperature

Let us find out what information is given to us.

m = 100 g

The temperature change can be calculated as,

ΔT = final temperature - initial temperature

ΔT = 50°C - 25°C = 25°C

C = 4.18 J/g-°C

Let us plug in the above values to find q.

q = 10450 J

Let us convert this to kJ

The amount of heat transferred was 10.45 kJ

23.5

23.5

47.0

i'm pretty sure it is letter c.) car