If 1600 cal of heat are added to 50 g of water initially at a temperature of 10 celsius what is the final temperature of the water?

Answer 1

To find the final temperature of the water, we can use the formula ( Q = mc\Delta T ), where ( Q ) is the heat added, ( m ) is the mass of the water, ( c ) is the specific heat capacity of water (approximately 4.18 J/g°C), and ( \Delta T ) is the change in temperature.

First, convert 1600 calories to joules: ( 1600 , \text{cal} \times 4.184 , \text{J/cal} = 6694.4 , \text{J} ). Then, using ( m = 50 , \text{g} ), we set up the equation:

[ 6694.4 , \text{J} = 50 , \text{g} \times 4.18 , \text{J/g°C} \times \Delta T ]

Solving for ( \Delta T ) gives ( \Delta T \approx 32.0 , \text{°C} ). Thus, the final temperature is ( 10 , \text{°C} + 32.0 , \text{°C} = 42.0 , \text{°C} ).