Question

how much heat is required to heat 50.0 g of gold from 25.0 to 29.0 °c if the specific heat is 0.0305 cal/g°c?
what is the final answer?
give two correct answers
select 2 correct answer(s)

1 g °c 1 1
(----------------)(----------------)(----------------)=0.16 cal

0.0305 cal 50.0 g 4.0 °c

0.0305 cal 50.0 g 4.0 °c
(--------)(--------)(--------)=6.1 cal

1 g °c 1 1
(0.0305 cal/1 g °c) (50.0 g) (4.0 °c)=6.1 cal
(1 g °c/0.0305 cal) (1/50.0 g) (1/4.0 °c)=0.16 cal

Explanation:

Step1: Identify the formula

The formula for heat $Q = mc\Delta T$, where $m$ is mass, $c$ is specific - heat, and $\Delta T$ is change in temperature.

Step2: Calculate the change in temperature

$\Delta T=T_2 - T_1=29.0^{\circ}C - 25.0^{\circ}C = 4.0^{\circ}C$.

Step3: Substitute values into the formula

Given $m = 50.0\ g$, $c=0.0305\ cal/g^{\circ}C$, and $\Delta T = 4.0^{\circ}C$. Then $Q=(50.0\ g)\times(0.0305\ cal/g^{\circ}C)\times(4.0^{\circ}C)$.

Step4: Perform the calculation

$Q=(50.0)\times(0.0305)\times(4.0)=6.1\ cal$.

Answer:

$(0.0305\ cal/1\ g^{\circ}C)\ (50.0\ g)\ (4.0^{\circ}C)=6.1\ cal$; $(50.0\ g)\ (0.0305\ cal/g^{\circ}C)\ (4.0^{\circ}C)=6.1\ cal$