Will maynez burns a 0.8-g peanut beneath 76 g of water, which increases in temperature from 22âc to 46 âc. (the specific heat capacity of water is 1.0 cal/(gââ

Based on the information provided, it appears that you will need to calculate the amount of heat absorbed by the water from the peanut that was burned. We are given the following information:

specific heat capacity, c = 1.0 cal/g°C
mass of water = 76 g
Ti = 22°C
Tf = 46°C
change in temperature, ΔT = 24°C

We can use the formula q = mcΔT to measure the amount of energy absorbed by the water to increase in tempature:

q = (76 g)(1.0 cal/g°C)(24°C)

q = 1824 cal

Therefore, the water absorbed 1824 calories from the peanut that was burned.

It would be (at) + cu (s)