Question

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 1

I belive it is 10.45, i have the same question

Answer 2

Answer : 10.45 kJ of heat was transferred.

Explanation :

The equation we need to use here is given as,

$q= m\times\ C \times \bigtriangleup T$

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.

$q = 100 g\times 4.18 \frac{J}{g ^{o}C} \times 25^{o} C$

q = 10450 J

Let us convert this to kJ

$10450 J \times \frac{1kJ}{1000J} = 10.45 kJ$

The amount of heat transferred was 10.45 kJ