John likes to drink tea (specific heat capacity for tea is 4100 J/kg°C). He makes himself a nice cup of tea with a volume of around 100 mL. He wants to drink his tea, but it is too far hot (99°C). How much cold water does he need to add if he wants the temperature of his tea to be around 40°C?
PLEASE GIVE ME ALL STEPS!!
PLEASE GIVE ME ALL STEPS!!
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YOU WILL DO THE MATH!!
With all the approximations and not giving the temp of the cold water, I could just guess: "around 50 mL". But that wouldn't be any fun.
The heat lost, QL, by the tea in cooling to 40°C is equal to the heat gained, QG, by the water in heating up to 40°C. That lost heat had to go someplace.
QL = Vt*sph_t*deltaT
where the volume of the tea is 100 mL, sph_t is 4100 J/kg°C , deltaT is 99°C-40°C
QG = Vw*spg_w*deltaT
where the volume of the water is the unknown, sph_w is 4186 J/kg°C, I'm going to say the cold water is at 10°C so deltaT for the water is 40°C-10°C.
Vt*sph_t*deltaT = Vw*spg_w*deltaT
Plug in the data and solve for Vw.
With all the approximations and not giving the temp of the cold water, I could just guess: "around 50 mL". But that wouldn't be any fun.
The heat lost, QL, by the tea in cooling to 40°C is equal to the heat gained, QG, by the water in heating up to 40°C. That lost heat had to go someplace.
QL = Vt*sph_t*deltaT
where the volume of the tea is 100 mL, sph_t is 4100 J/kg°C , deltaT is 99°C-40°C
QG = Vw*spg_w*deltaT
where the volume of the water is the unknown, sph_w is 4186 J/kg°C, I'm going to say the cold water is at 10°C so deltaT for the water is 40°C-10°C.
Vt*sph_t*deltaT = Vw*spg_w*deltaT
Plug in the data and solve for Vw.