Question

What is the final temperature after 840 calories is absorbed by 10.0 g of water at 25.0 °C? The specific heat of water is 1 calorie/gram °C.

A) 30.0 °C

B) 32.0 °C

C) 35.0 °C

D) 40.0 °C

Answer 1

To calculate the final temperature after absorbing 840 calories, we can use the equation q = mcΔT. Plugging in the values, we find that the final temperature is 109 °C.

Step-by-step explanation:

To calculate the final temperature after 840 calories is absorbed by 10.0 g of water at 25.0 °C, we can use the equation:

q = mcΔT

Where:
q is the heat absorbed by the water (in calories)
m is the mass of the water (in grams)
c is the specific heat of water (1 calorie/gram °C)
ΔT is the change in temperature (final temperature - initial temperature) in °C

Plugging in the values, we get:
q = (10.0 g)(1 calorie/gram °C)(final temperature - 25.0 °C) = 840 calories

Simplifying the equation, we have:
final temperature - 25.0 °C = 840 calories / (10.0 g)(1 calorie/gram °C)
final temperature - 25.0 °C = 84 °C/g
final temperature = 25.0 °C + 84 °C/g

Calculating the final temperature, we find that the answer is 109 °C.