Question

You add 1,590 cal of heat energy to 379 g of water, initially at 11 ◦c. the specific heat capacity of water is 1.0 cal/g◦c. what is the water’s final temperature?

Answer 1

The water's final temperature after adding 1,590 cal of heat to 379 g of water initially at 11°C is approximately 15.2°C.

Step-by-step explanation:

The final temperature of the water can be calculated using 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, c is given as 1.0 cal/g°C, which makes our calculation straightforward as we won't need to convert to joules. Plugging in the values, we have:

1,590 cal = 379 g × 1.0 cal/g°C × ΔT

ΔT (change in temperature) = 1,590 cal / (379 g × 1.0 cal/g°C) = 4.2°C (approximately)

The final temperature of the water is the initial temperature plus the change in temperature: 11°C + 4.2°C = 15.2°C.