Question

A scientist heated a tank containing 50 g of water. The specific heat of water is 4.18 J/gºC. The temperature of the water increased from 25ºC to 37ºC. How much heat energy did the water absorb?

Your answer:


2,508 Joules

-2,508 Joules

5,225 Joules

7,733 Joules

Answer 1

The water absorbed 2,508 Joules of heat energy.

Step-by-step explanation:

The specific heat of water is 4.18 J/gºC. To calculate the amount of heat energy absorbed, we can use the formula q = mcΔT, where q is the heat energy, 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 50 g and the change in temperature is 12°C (37°C - 25°C).

Plugging in these values, we get:

q = (50 g)(4.18 J/gºC)(12ºC) = 2,508 Joules

Answer 2

I got 5,225 by 50x4.18= 209(25)=5,225