If 41grams of water at 24°c absorbs 2208 j of heat energy, what will be the final temperature of the water?

The problem can solved using the heat equation which is expressed as:

H = mCΔT

 where H is the energy absorbed or released, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature.

2208 J = 41 g x 4.18 J/g·°C x ( T - 24 °C) 
T = 36.88 °C

It’s c it stays the same. and the atomic number goes up by 1. a neutron has changed into a proton plus an electron. the electron is emitted as a beta particle.

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