# A cup of tea containing 400.0 g of liquid at 20.0 degree C was heated. If 58.5 kJ of heat was added to the tea, what would be the final temperature? Assume that the specific heat is the same as that of water( 4.18 J g c)

Mar 31, 2015

The final temperature of the tea is 55.0 °C.

The formula to use is

q = mcΔT,

where $q$ is the heat involved, $m$ is the mass, $c$ is the specific heat capacity, and ΔT is the temperature change.

$q = \text{58 500 J}$
$m = \text{400.0 g}$
$c = \text{4.18 J·g⁻¹°C⁻¹}$
ΔT = "?"

Solve the equation for ΔT.

ΔT = q/(mc) = ("58 500" cancel("J"))/(400.0 cancel("g") × 4.18 cancel("J·g⁻¹")"°C⁻¹") = "35.0 °C"

Δ T = T_2 – T_1

T_2 = T_1 + ΔT = "20.0 °C + 35.0 °C" = "55.0 °C"