Question

In setting up an aquarium, the heater transfers 1,200,000 J of heat to 75,000 g of water. What is the increase in the water’s temperature if the specific heat of water is 4.18 J/g*˚C?

Answer 1

The increase in the water's temperature is approximately 3.84°C when 1,200,000 J of heat is transferred to 75,000 g of water with a specific heat capacity of 4.18 J/g°C.

Step-by-step explanation:

To calculate the increase in water's temperature when 1,200,000 J of heat is transferred to 75,000 g of water, we use the formula: Q = mcΔT, where Q is the heat added, m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature. Here, m = 75,000 g, c = 4.18 J/g°C, and Q = 1,200,000 J.

Solving for ΔT, we get: ΔT = Q/(mc)

ΔT = 1,200,000 J / (75,000 g × 4.18 J/g°C)

ΔT = 3.84°C

So the increase in the water's temperature is approximately 3.84°C.