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.

a) 10.5 J
b) 10.5 kJ
c) 100 J
d) 100 kJ

Answer 1

To calculate the amount of heat transferred, use the equation Q=mcΔT. In this case, the answer is 10.5 J.

Step-by-step explanation:

To calculate the amount of heat transferred, we can use the equation Q=mcΔT, where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

In this case, the mass of the water is 100 g and the change in temperature is 25°C - 50°C = -25°C. The specific heat capacity of water is 4.18 J/g-°C.

Plugging in the values, we have Q = 100 g x 4.18 J/g-°C x -25°C = -10,450 J. Since heat is transferred from the water, the negative sign indicates heat lost. Therefore, the correct answer is option a) 10.5 J.